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On uniqueness of elastic scattering from a cavity

Tianjiao Wang Yiwen Lin Xiang Xu
Abstract

The paper considers direct and inverse elastic scattering from a cavity in homogeneous medium with Dirichlet and Neumann boundary conditions. For direct scattering, existence and uniqueness are derived by variation approach. For inverse scattering, Fre´\acute{\rm e}chet derivatives of the solution operators are investigated, which give local stability for Dirichlet case.

keywords:
elastic cavity , variation formulation , Navier equations , existence , uniqueness
MSC:
35P25 , 35A15 , 35R30
journal: Inverse Problems and Imaging
\affiliation

[inst1]organization=School of Mathematical Sciences,addressline=Zhejiang University, city=Hangzhou, postcode=310058, country=P. R. China

\affiliation

[inst2]organization=School of Mathematical Sciences and Institute of Natural Sciences,addressline=Shanghai Jiao Tong University, city=Shanghai, postcode=200240, country=P. R. China

inst3inst3footnotetext: Corresponding author, [email protected]. This work was supported in part by National Natural Science Foundation of China (11621101, 12071430, 12201404), Key Laboratory of Collaborative Sensing and Autonomous Unmanned Systems of Zhejiang Province and Postdoctoral Science Foundation of China (2021TQ0203).

1 Introduction

In this paper, we consider a time-harmonic plane elastic wave incident on a cavity which can be regarded as a local perturbation below the plane with wide-ranging applications as engine inlet ducts, cracks, gaps and so on.

For electromagnetic incident waves, there are considerable research results in literature. For instance, Ammari, Bao and Wood proved uniqueness and existence of electromagnetic cavity problems respectively to TE and TM polarization by variational approaches in [1], and by integral equation methods in [2]. They also extend similar results for Maxwell’s equations in [3] by variational approaches. Recently, Bao et al established a stability result explicitly dependent on wave number for TE polarization based on Fourier analysis in energy space in [4] and TM polarization in [5]. Numerical methods for large cavities can be seen in [6, 7, 8].

Compared to electromagnetic scattering, elastic scattering also has wide applications in seismology and geophysics such as [9, 10, 11] and attracts more attention recently. However, elastic scattering problems have not been studied intensively due to their inherent difficulties arising from the governing equations and boundary conditions. For periodic surface, Arens studied quasi-periodic Green tensor and elastic potentials in [12] and used Rayleigh expansions and Rellich identities to prove uniqueness and integral equation methods to prove existence in [13]. Elschner and Hu applied variational approaches to get similar results in [14]. For general unbounded rough surfaces, Arens studied Green tensor, elastic potentials and upward radiation condition (UPRC), and proved uniqueness by integral equation methods and existence by integral equation methods, see [15, 16, 17]. Elschner and Hu [18] gave an equivalent form of UPRC and proved uniqueness and existence by variational approaches. They [19] also considered solvability in weighted Sobolev spaces and proved existence and uniqueness of elastic scattering from unbounded rough surface with an incident wave. Moreover, Hu, Yuan and Zhao [20] combined variation formula and integral equation to prove uniqueness and existence of two-dimensions local perturbed surface scattering problem with Lipschitz graph. For cavities, Hu, Li and Zhao considered elastic scattering in three-dimensions and proved the uniqueness in [21] for Dirichlet boundary conditions.

Inverse cavity problem is to reconstruct the shape of a cavity by measured far-field data on the artificial boundary. In general, to reconstruct a shape, Fre´\acute{\rm e}chet derivative is commonly used. For instance, Bao, Gao and Li considered TE and TM polarization in [22] and proved uniqueness for lossy medium and studied the domain derivative and local stability for inverse electromagnetic cavity. A similar results for Maxwell’s equations was extended in [23]. Bao and Lai [24] proposed an optimization scheme to design a cavity for radar cross section reduction. Recently, Hu, Yuan, Zhao considered domain derivative for the inverse elastic scattering by a locally perturbed rough surface in [20]. To the authors’ best knowledge, research on inverse elastic scattering from cavities is still at its infant stage.

This paper intends to extend electromagnetic scattering from a cavity to elastic scattering. By employing transparent boundary conditions given by Dirichlet to Neumann (DtN) and Neumann to Dirichlet (NtD) operators, the boundary value problems can be reduced into a bounded domain so that the Fredholm alternative can be applied. Then, the variational approach is used to prove existence and uniqueness of the elastic scattering problem in two-dimensions with Dirichlet condition or Neumann condition. For Dirichlet problem, the existence in two-dimensions is actually trivial compared to three dimensions in [21]. However, we can obtain the uniqueness for two-dimensions while it is still open for three dimensions. Neumann problem is similar to Dirichlet problem with additional difficulties for the NtD operator. On inverse elastic cavity problem, this paper applies the method of changing variable to give the Fre´\acute{\rm e}chet derivative for Dirichlet problem and Neumann problem, respectively. The difficulty for inverse Neumann problem lies in the boundary value problem reduced by it in Section 2 has inhomogeneous Neumann boundary condition by directly applying NtD operator. Hence, in order to obtain the TBC deduced by the DtN operator, we use an artificial boundary to rewrite the boundary value problem for Neumann case.

The paper is outlined as follows. In Section 2, the direct problems are reduced into variation problems in a bounded domain by DtN operator and NtD operator. In Section 3 and Section 4, existence and uniqueness for Dirichlet problem and Neumann problem are examined respectively. Moreover, the solutions to variation problems can be extended to satisfy Kurpradze-Sommerfeld radiation condition and Navier equations in upper half-space. In section 5, the Fre´\acute{\rm e}chet derivatives for Dirichlet problem and Neumann problem are given respectively and a local stability result is given for Dirichlet problem. Conclusion is given in Section 6.

2 Problem formulation

Consider a time-harmonic plane wave incident on a cavity, which is shown in Figure 1, denoted by DD with boundary ΓS\Gamma\cup S, where Γ{x2=0}\Gamma\subset\{x_{2}=0\} is bounded and SS is assumed Lipschitz continuous. It is noted that SS is not necessary a graph of some function. In addition, denote Γc={x2=0}\Γ\Gamma^{c}=\{x_{2}=0\}\backslash\Gamma. The medium is assumed to be homogeneous.

Refer to caption
Figure 1: The problem geometry

Suppose the incident time-harmonic plane wave takes the form

ui=cpdexp(ikp(x1sinθx2cosθ))+csdexp(iks(x1sinθx2cosθ)),u_{i}=c_{p}\vec{d}\,\text{exp}(ik_{p}(x_{1}\sin{\theta}-x_{2}\cos{\theta}))+c_{s}\vec{d}^{\bot}\text{exp}(ik_{s}(x_{1}\sin{\theta}-x_{2}\cos{\theta})),

where

cp,cs,d=(sinθ,cosθ),d=(cosθ,sinθ),θ(π/2,π/2).c_{p},c_{s}\in\mathbb{C},\quad\vec{d}=(\sin{\theta},-\cos{\theta}),\quad\vec{d}^{\bot}=(\cos{\theta},\sin{\theta}),\quad\theta\in(-\pi/2,\pi/2).

Define the compressional and shear wave numbers by

kp:=ω/2μ+λ,ks:=ω/μ.k_{p}:=\omega/\sqrt{2\mu+\lambda},\quad k_{s}:=\omega/\sqrt{\mu}.

For simplicity, throughout this paper, it assumes cp=kpc_{p}=k_{p}, cs=0c_{s}=0. Consider total wave field uu which satisfies the following Navier equations with Lame´\acute{\rm e} constants λ>0\lambda>0, μ>0\mu>0 and frequency ω>0\omega>0

μΔu+(μ+λ)(u)+ω2u=0inD{x2>0}.\displaystyle\mu\Delta u+(\mu+\lambda)\nabla(\nabla\cdot u)+\omega^{2}u=0\quad{\rm in}\quad D\cup\{x_{2}>0\}. (2.1)

For convenience, denote by Δu=μΔu+(μ+λ)(u)\Delta^{*}u=\mu\Delta u+(\mu+\lambda)\nabla(\nabla\cdot u). This paper considers two kinds of boundary conditions, i.e., Dirichlet condition

u=0onSΓc,u=0\quad{\rm on}\quad S\cup\Gamma^{c}, (2.2)

and Neumann condition

Tu=0onSΓc,Tu=0\quad\text{on}\quad S\cup\Gamma^{c}, (2.3)

where the differential operator is defined Tu:=μnu+(λ+μ)ndivuTu:=\mu\partial_{n}u+(\lambda+\mu)\vec{n}{\rm div}\,u. Consider the Helmholtz decomposition for uu, i.e.,

u=i(gradϕ+curlψ),u=-i({\rm grad}\,\phi+{\overrightarrow{\rm curl}}\,\psi), (2.4)

with

ϕ=ikp2divu, and ψ=iks2curlu,\quad\phi=-\frac{i}{k^{2}_{p}}{\rm div}\,u,\text{ and }\psi=\frac{i}{k^{2}_{s}}{\rm curl}\,u, (2.5)

where curlu=(2u1,1u2)\overrightarrow{\text{curl}}\,u=(\partial_{2}u_{1},-\partial_{1}u_{2})^{\top} and curlu=1u22u1\text{curl}\,u=\partial_{1}u_{2}-\partial_{2}u_{1}. Then, ϕ\phi and ψ\psi satisfy the homogeneous Helmholtz equations

(Δ+kp2)ϕ=0and(Δ+ks2)ψ=0,x2>0.(\Delta+k^{2}_{p})\phi=0\quad{\rm and}\quad(\Delta+k^{2}_{s})\psi=0,\quad x_{2}>0. (2.6)

The total wave field consists of incident wave uiu_{i}, reflected wave uru_{r} and scattering wave usu_{s}, i.e., u=ui+ur+usu=u_{i}+u_{r}+u_{s}. Here, the reflected wave uru_{r} satisfies

Δur+ω2ur=0in{x2>0},\displaystyle\Delta^{*}u_{r}+\omega^{2}u_{r}=0\quad{\rm in}\quad\{x_{2}>0\},
ur+ui=0on{x2=0},\displaystyle u_{r}+u_{i}=0\quad\text{on}\quad\{x_{2}=0\}, (2.7)
orT(ui+ur)=0on{x2=0}.\displaystyle\text{or}\quad T(u_{i}+u_{r})=0\quad\text{on}\quad\{x_{2}=0\}. (2.8)

It is easy to know that the reflected wave exists and can be expressed by the following functions

u1\displaystyle u_{1} =(α,β)exp(i(αx1βx2)),\displaystyle=\left(\alpha,\beta\right)^{\top}\text{exp}(i(\alpha x_{1}-\beta x_{2})),
u2\displaystyle u_{2} =(α,β)exp(i(αx1+βx2)),\displaystyle=\left(\alpha,-\beta\right)^{\top}\text{exp}(i(\alpha x_{1}+\beta x_{2})),
u3\displaystyle u_{3} =(η,α)exp(i(αx1+ηx2)),\displaystyle=\left(\eta,-\alpha\right)^{\top}\text{exp}(i(\alpha x_{1}+\eta x_{2})),

with α=kpsinθ,β=kpcosθ\alpha=k_{p}\sin{\theta},\,\beta=k_{p}\cos{\theta} and η=ks2α2\eta=\sqrt{k_{s}^{2}-\alpha^{2}} which are linearly independent solutions to Navier equations. Combing the Dirichlet boundary condition gives

ur,1=u1α2βηα2+βηu22αβα2+βηu3.u_{r,1}=u_{1}-\frac{\alpha^{2}-\beta\eta}{\alpha^{2}+\beta\eta}u_{2}-\frac{2\alpha\beta}{\alpha^{2}+\beta\eta}u_{3}.

And the Neumann boundary condition (2.8) yields

ur,2=\displaystyle u_{r,2}= (λ+μ)kp2+μβ2η+μα2β2μα2βu1\displaystyle\frac{(\lambda+\mu)k_{p}^{2}+\mu\beta^{2}\eta+\mu\alpha^{2}\beta}{2\mu\alpha^{2}\beta}u_{1}
+(λ+μ)kp2μβ2η+μα2β2μα2βu2+(μ+λ)kp2+μβ2μαηu3.\displaystyle+\frac{-(\lambda+\mu)k_{p}^{2}-\mu\beta^{2}\eta+\mu\alpha^{2}\beta}{2\mu\alpha^{2}\beta}u_{2}+\frac{(\mu+\lambda)k^{2}_{p}+\mu\beta^{2}}{\mu\alpha\eta}u_{3}.

For the scattering wave usu_{s}, it is required to satisfy the following half-plane Kupradze-Sommerfeld radiation condition,

limr0r1/2(rϕ(x)ikpϕ(x))=0in+,\lim_{r\to 0}r^{1/2}(\partial_{r}\phi(x)-ik_{p}\phi(x))=0\quad{\rm in}\quad\mathbb{R}^{+}, (2.9)

and

limr0r1/2(rψ(x)iksψ(x))=0in+,\lim_{r\to 0}r^{1/2}(\partial_{r}\psi(x)-ik_{s}\psi(x))=0\quad{\rm in}\quad\mathbb{R}^{+}, (2.10)

with r=|x|r=|x| for ϕ=ikp2divus\phi=-\frac{i}{k^{2}_{p}}{\rm div}u_{s}, ψ=iks2curlus\psi=\frac{i}{k^{2}_{s}}{\rm curl}u_{s}.

Now we can state the following two problems respectively.

Dirichlet problem: Find uu which satisfies (2.1)-(2.2), where usu_{s} satisfies Kupradze-Sommerfeld radiation condition (2.9)-(2.10).

Neumann problem: Find uu which satisfies (2.1) and (2.3), where usu_{s} satisfies Kupradze-Sommerfeld radiation condition (2.9)-(2.10).

To investigate the above two problems in an unbounded domain, it is necessary to introduce transparent boundary conditions (TBC) to reduce the model problems into a bounded domain.

By Helmholtz decomposition, we have usu_{s} satisfying (2.4) with corresponding ϕ\phi, ψ\psi satisfying (2.5)-(2.6). Applying the Fourier transform to (2.6) with respect to x1x_{1} and using radiation condition gives

ϕ^=P(ξ)exp(ix2γp(ξ)),ψ^=S(ξ)exp(ix2γs(ξ)),\hat{\phi}=P(\xi)\,\text{exp}(ix_{2}\gamma_{p}(\xi)),\quad\hat{\psi}=S(\xi)\,\text{exp}(ix_{2}\gamma_{s}(\xi)),

with

γp(ξ)=kp2ξ2,γs(ξ)=ks2ξ2.\gamma_{p}(\xi)=\sqrt{k^{2}_{p}-\xi^{2}},\quad\gamma_{s}(\xi)=\sqrt{k^{2}_{s}-\xi^{2}}.

Denote by u^(ξ)=u(ξ)\hat{u}(\xi)=\mathcal{F}u(\xi) the Fourier transformation of uu with respect to x1x_{1}. Then the Fourier transformation of uu is given by

(u^s,1u^s.2)=(ξi2i2ξ)(P(ξ)exp(ix2γp)S(ξ)exp(ix2γs)).\left(\begin{array}[]{cc}\hat{u}_{s,1}\\ \hat{u}_{s.2}\end{array}\right)=\left(\begin{array}[]{cc}\xi&-i\partial_{2}\\ -i\partial_{2}&-\xi\end{array}\right)\left(\begin{array}[]{cc}P(\xi)\,\text{exp}(ix_{2}\gamma_{p})\\ S(\xi)\,\text{exp}(ix_{2}\gamma_{s})\end{array}\right). (2.11)

Let x2=0x_{2}=0,

u^s(ξ,0)=(u^s,1(ξ,0)u^s,2(ξ,0))=(ξγsγpξ)(P(ξ)S(ξ)),\hat{u}_{s}(\xi,0)=\left(\begin{array}[]{cc}\hat{u}_{s,1}(\xi,0)\\ \hat{u}_{s,2}(\xi,0)\\ \end{array}\right)=\left(\begin{array}[]{cc}\xi&\gamma_{s}\\ \gamma_{p}&-\xi\end{array}\right)\left(\begin{array}[]{cc}P(\xi)\\ S(\xi)\end{array}\right), (2.12)

which implies

(P(ξ)S(ξ))=1ξ2+γpγs(ξγsγpξ)(u^s,1(ξ,0)u^s,2(ξ,0)).\left(\begin{array}[]{cc}P(\xi)\\ S(\xi)\end{array}\right)=\frac{1}{\xi^{2}+\gamma_{p}\gamma_{s}}\left(\begin{array}[]{cc}\xi&\gamma_{s}\\ \gamma_{p}&-\xi\end{array}\right)\left(\begin{array}[]{cc}\hat{u}_{s,1}(\xi,0)\\ \hat{u}_{s,2}(\xi,0)\\ \end{array}\right). (2.13)

Inserting (2.13) into (2.11) arrives at the following representation for usu_{s}:

us=12π(exp(ix2γp(ξ))Mp(ξ)+exp(ix2γs(ξ))Ms(ξ))u^s(ξ,0)exp(ix1ξ)dξu_{s}=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}(\text{exp}(ix_{2}\gamma_{p}(\xi))M_{p}(\xi)+\text{exp}(ix_{2}\gamma_{s}(\xi))M_{s}(\xi))\hat{u}_{s}(\xi,0)\text{exp}(ix_{1}\xi)\,\mathrm{d}\xi (2.14)

in {x2>0}\{x_{2}>0\} with

Mp(ξ)=1ξ2+γpγs(ξ2γsξγpξγpγs),Ms(ξ)=1ξ2+γpγs(γpγsγsξγpξξ2).M_{p}(\xi)=\frac{1}{\xi^{2}+\gamma_{p}\gamma_{s}}\left(\begin{array}[]{cc}\xi^{2}&\gamma_{s}\xi\\ \gamma_{p}\xi&\gamma_{p}\gamma_{s}\end{array}\right),\quad M_{s}(\xi)=\frac{1}{\xi^{2}+\gamma_{p}\gamma_{s}}\left(\begin{array}[]{cc}\gamma_{p}\gamma_{s}&-\gamma_{s}\xi\\ -\gamma_{p}\xi&\xi^{2}\end{array}\right).

Recalling the definition of differential operator TT, we have

Tus:=μnus+(λ+μ)ndivusonΓ.Tu_{s}:=\mu\partial_{n}u_{s}+(\lambda+\mu)\vec{n}{\rm div}\,u_{s}\quad{\rm on}\quad\Gamma. (2.15)

Combing (2.14)-(2.15) and direct calculation implies that

Tus=12πM(ξ)u^s(ξ,0)exp(ixξ)dξ,Tu_{s}=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}M(\xi)\hat{u}_{s}(\xi,0)\text{exp}(ix\xi)\,\mathrm{d}\xi, (2.16)

where

M(ξ)=iξ2+γpγs(ω2γpξω2+ξμ(ξ2+γpγs)ξω2ξμ(ξ2+γpγs)ω2γs).M(\xi)=\frac{i}{\xi^{2}+\gamma_{p}\gamma_{s}}\left(\begin{array}[]{cc}\omega^{2}\gamma_{p}&-\xi\omega^{2}+\xi\mu(\xi^{2}+\gamma_{p}\gamma_{s})\\ \xi\omega^{2}-\xi\mu(\xi^{2}+\gamma_{p}\gamma_{s})&\omega^{2}\gamma_{s}\end{array}\right). (2.17)

Then the Dirichlet to Neumann operator can be defined by

𝒯f:=1(Mf^),fH1/2()2.\mathcal{T}f:=\mathcal{F}^{-1}(M\hat{f}),\quad f\in H^{1/2}(\mathbb{R})^{2}.

So Tus=𝒯usTu_{s}=\mathcal{T}u_{s} on Γ\Gamma. Noting that ui+ur=0u_{i}+u_{r}=0 on {x2=0}\{x_{2}=0\} and u=0u=0 on Γc\Gamma^{c}, then TBC for Dirichlet problem can be given by

Tu=𝒯u+gonΓ,withg=T(ui+ur).Tu=\mathcal{T}u+g\quad{\rm on}\quad\Gamma,\quad\text{with}\quad g=T(u_{i}+u_{r}).

By TBC, Dirichlet problem can be reformulated as

Δu+ω2u=0inD,\displaystyle\Delta^{*}u+\omega^{2}u=0\quad{\rm in}\quad D,
u=0onS,\displaystyle u=0\quad{\rm on}\quad S,
Tu=𝒯u+gonΓ.\displaystyle Tu=\mathcal{T}u+g\quad{\rm on}\quad\Gamma.

Denote V=HS1(D)2={uH1(D)2:u=0onS}V=H^{1}_{S}(D)^{2}=\{u\in H^{1}(D)^{2}:u=0\,\mathrm{on}\,S\}. Then for u,vVu,\,v\in V, the Betti formula gives

0=D(Δ+ω2)uv¯dx=D(u,v¯)ω2uv¯dxΓ(𝒯u+g)v¯ds,0=-\int_{D}(\Delta^{*}+\omega^{2})u\cdot\bar{v}\,\mathrm{d}x=\int_{D}\mathcal{E}(u,\bar{v})-\omega^{2}u\cdot\bar{v}\,\mathrm{d}x-\int_{\Gamma}(\mathcal{T}u+g)\cdot\bar{v}\,\mathrm{d}s,

where

(u,v)=μ(u1v1+u2v2)+(λ+μ)(u)(v).\mathcal{E}(u,v)=\mu(\nabla u_{1}\cdot\nabla v_{1}+\nabla u_{2}\cdot\nabla v_{2})+(\lambda+\mu)(\nabla\cdot u)(\nabla\cdot v). (2.18)

Define a sesquilinear form B:V×VB\,:V\times V\to\mathbb{C} as

B(u,v)=D(u,v¯)ω2uv¯dxΓ𝒯uv¯ds.B(u,v)=\int_{D}\mathcal{E}(u,\bar{v})-\omega^{2}u\cdot\bar{v}\,\mathrm{d}x-\int_{\Gamma}\mathcal{T}u\cdot\bar{v}\,\mathrm{d}s.

Now we obtain variation formula of Dirichlet boundary value problem.

Variation problem 1: find uVu\in V such that B(u,v)=(g,v)Γ,vVB(u,v)=(g,v)_{\Gamma},\,\forall v\in V.

It is similar to give TBC and variation formula corresponding to Neumann problem. Note that in this case, it is not usu_{s} but TusTu_{s}, which has compact support on Γ\Gamma. So u^s\hat{u}_{s} may not be integrable on {x2=0}\{x_{2}=0\}. Thus, (2.13) and (2.16) may not hold in this case. However, usu_{s} can be represented by Tus^(ξ,0)\widehat{Tu_{s}}(\xi,0) similarly. According to (2.14), we obtain

Tus^(ξ,x2)=(μ20i(λ+μ)ξ(λ+2μ)2)(u^s,1u^s,2).\widehat{Tu_{s}}(\xi,x_{2})=\left(\begin{array}[]{cc}\mu\partial_{2}&0\\ i(\lambda+\mu)\xi&(\lambda+2\mu)\partial_{2}\end{array}\right)\left(\begin{array}[]{cc}\hat{u}_{s,1}\\ \hat{u}_{s,2}\end{array}\right). (2.19)

Combing (2.11) and (2.19) yields

Tus^(ξ,x2)=(μ20i(λ+μ)ξ(λ+2μ)y)(ξi2i2ξ)(P(ξ)exp(ix2γp)S(ξ)exp(ix2γs)).\widehat{Tu_{s}}(\xi,x_{2})=\left(\begin{array}[]{cc}\mu\partial_{2}&0\\ i(\lambda+\mu)\xi&(\lambda+2\mu)\partial_{y}\end{array}\right)\left(\begin{array}[]{cc}\xi&-i\partial_{2}\\ -i\partial_{2}&-\xi\end{array}\right)\left(\begin{array}[]{cc}P(\xi)\,\text{exp}(ix_{2}\gamma_{p})\\ S(\xi)\,\text{exp}(ix_{2}\gamma_{s})\end{array}\right).

Take x2=0x_{2}=0, then P,SP,\,S can be represented by Tus^(ξ,0)\widehat{Tu_{s}}(\xi,0). Insert the representation in (2.11), we have

u^s(ξ,x2)=(exp(ix2γp(ξ))Mp(ξ)+exp(ix2γs(ξ))Ms(ξ))M1Tus^(ξ,0).\hat{u}_{s}(\xi,x_{2})=\Big{(}\text{exp}\big{(}ix_{2}\gamma_{p}(\xi)\big{)}M_{p}(\xi)+\text{exp}\big{(}ix_{2}\gamma_{s}(\xi)\big{)}M_{s}(\xi)\Big{)}M^{-1}\widehat{Tu_{s}}(\xi,0).

Taking inverse Fourier transformation and x2=0x_{2}=0 gives

us(ξ,0)=1(M1us^(ξ,0)).u_{s}(\xi,0)=\mathcal{F}^{-1}(M^{-1}\widehat{u_{s}}(\xi,0)).

Hence NtD operator Λ\varLambda can be defined by

Λ(f):=1(M1f^),fH1/2()2,\varLambda(f):=\mathcal{F}^{-1}(M^{-1}\hat{f}),\quad f\in H^{-1/2}(\mathbb{R})^{2},

which implies

Λ(Tus)=usonΓ.\varLambda(Tu_{s})=u_{s}\quad\text{on}\quad\Gamma.

Denote h=ur+uih=u_{r}+u_{i} and g=(Δ+ω2)hg=-(\Delta^{*}+\omega^{2})h. Then the Neumann problem can be reformulated as

Δus+ω2us=0inD,\displaystyle\Delta^{*}u_{s}+\omega^{2}u_{s}=0\quad{\rm in}\quad D,
Tus=ThonS,\displaystyle Tu_{s}=-Th\quad{\rm on}\quad S, (2.20)
us=Λ(Tus)onΓ.\displaystyle u_{s}=\varLambda(Tu_{s})\quad{\rm on}\quad\Gamma.

Define the function space

W={uH1(D)2:TuH01/2(Γ)2,u=Λ(Tu)onΓ},W=\{u\in H^{1}(D)^{2}:Tu\in H^{-1/2}_{0}(\Gamma)^{2},u=\varLambda(Tu)\,\text{on}\,\Gamma\},

where

H01/2(Γ)2={uH1/2(Γ)2:Λ(u~)H1/2(K)2,K2},H^{-1/2}_{0}(\Gamma)^{2}=\{u\in H^{-1/2}(\Gamma)^{2}:\varLambda(\tilde{u})\in H^{1/2}(K)^{2},\,{\forall}K\subset\subset\mathbb{R}^{2}\},

and u~\tilde{u} is zero extension of uu. In fact, it is clear that H01/2(Γ)2H^{-1/2}_{0}(\Gamma)^{2} is the dual space of H1/2(Γ)2H^{1/2}(\Gamma)^{2} in [17]. The sesquilinear form \mathcal{B}: W×WW\times W\rightarrow\mathbb{C} is defined by

(v,w)=D(v,w¯)ω2vw¯dxΓTuΛ(Tw)¯ds.\mathcal{B}(v,w)=\int_{D}\mathcal{E}(v,\bar{w})-\omega^{2}v\cdot\bar{w}\,\mathrm{d}x-\int_{\Gamma}Tu\cdot\overline{\varLambda(Tw)}\,\mathrm{d}s.

Now we can give the following variation formula by the Betti formula.

Variation problem 2: find usWu_{s}\in W such that

(us,w)=SThw¯dsDgw¯dx,wW.\mathcal{B}(u_{s},w)=\int_{S}-Th\cdot\bar{w}\,\text{d}s-\int_{D}g\cdot\bar{w}\,\text{d}x,\quad\forall w\in W.

Next two sections consider existence and uniqueness of solutions to the above two variation problems. Moreover, they can be extended to {x2>0}\{x_{2}>0\} satisfying Kurpradze-Sommerfeld radiation condition.

3 Existence and uniqueness for Dirichlet problem

Since this problem is in a bounded domain with Lipschitz continuous boundary, we have the well-known compact embedding result that H1(D)H^{1}(D) is a compact subset of L2(D)L^{2}(D). Hence uniqueness of variation problem and Fredholm alternative yield existence directly. To this end, we show that the continuity of DtN operator results in the continuity of the sesquilinear form BB, which is stated in the following lemma with the proof omitted.

Lemma 3.1.

The DtN operator 𝒯\mathcal{T} is a continuous linear operator from H01/2(Γ)2H^{1/2}_{0}(\Gamma)^{2} to H1/2(Γ)2H^{-1/2}(\Gamma)^{2} .

In order to use Fredholm alternative, it needs to show the sesquilinear form satisfies the Gå\mathring{\rm a}rding inequality. Take real part of BB,

B(u,u)=D(u,u¯)ω2uu¯dxΓ𝒯uu¯ds.\Re\,B(u,u)=\int_{D}\mathcal{E}(u,\bar{u})-\omega^{2}u\cdot\bar{u}\,\mathrm{d}x-\Re\int_{\Gamma}\mathcal{T}u\cdot\bar{u}\,\mathrm{d}s.

Considering (2.18) gives

D(u,u¯)dx=μuL2(D)22+(λ+μ)uL2(D)22μuL2(D)22.\int_{D}\mathcal{E}(u,\bar{u})\,\mathrm{d}x=\mu\|\nabla u\|^{2}_{L^{2}(D)^{2}}+(\lambda+\mu)\|\nabla\cdot u\|^{2}_{L^{2}(D)^{2}}\geq\mu\|\nabla u\|^{2}_{L^{2}(D)^{2}}.

It turns out that

B(u,u)μuL2(D)22ω2uL2(D)22𝒯u,uΓ.\Re\,B(u,u)\geq\mu\|\nabla u\|^{2}_{L^{2}(D)^{2}}-\omega^{2}\|u\|^{2}_{L^{2}(D)^{2}}-\Re\,\left\langle\mathcal{T}u,u\right\rangle_{\Gamma}. (3.1)

By Plancherel identity,

𝒯u,uΓ=Mu,uΓ,-\Re\,\left\langle\mathcal{T}u,u\right\rangle_{\Gamma}=-\left\langle\Re M\,\mathcal{F}u,\mathcal{F}u\right\rangle_{\Gamma},

where M=(M+M¯)/2\Re\,M=(M+\bar{M}^{\top})/2. However, the sign of 𝒯u,uΓ-\Re\,\left\langle\mathcal{T}u,u\right\rangle_{\Gamma} is undeterminied because it lacks the result that M(ξ)-\Re\,M(\xi) is positive definite or negative definite. Fortunately, for sufficiently large |ξ||\xi|, we know M(ξ)-\Re\,M(\xi) is positive definite, given in Lemma 3.2.

Lemma 3.2.

M(ξ)>0-\Re\,M(\xi)>0 for |ξ|>ks|\xi|>k_{s}.

The proof is also omitted here as it is trivial compared as the results in [18]. Noting that

u,u=|ξ|>ksM(ξ)|u^|2dξ+|ξ|ksM(ξ)|u^|2dξ,\left\langle\mathcal{F}u,\mathcal{F}u\right\rangle=\int_{|\xi|>k_{s}}M(\xi)|\hat{u}|^{2}\mathrm{d}\,\xi+\int_{|\xi|\leq k_{s}}M(\xi)|\hat{u}|^{2}\mathrm{d}\,\xi,

and using Lemma 3.2, we have

u,u|ξ|ksM(ξ)|u^|2dξ.-\left\langle\Re\,\mathcal{F}u,\mathcal{F}u\right\rangle\geq-\int_{|\xi|\leq k_{s}}\Re\,M(\xi)|\hat{u}|^{2}\mathrm{d}\,\xi.

Therefore, it suffices to estimate the intergal in {|ξ|ks}\{|\xi|\leq k_{s}\}. The following inequality is similar as the three-dimensions case in [20].

Lemma 3.3.

For any uVu\in V, we have B(u,u)>C1uL2(D)22C2uL2(D)22\Re\,B(u,u)>C_{1}\|\nabla u\|^{2}_{L^{2}(D)^{2}}-C_{2}\|u\|^{2}_{L^{2}(D)^{2}}, where C1,C2C_{1},\,C_{2} are positive constants which depend on λ,μ,ω\lambda,\,\mu,\,\omega.

Next the uniqueness and existence of the variation problem can be derived by Lemma 3.3 and the Fredholm alternative.

Theorem 3.1.

Variation problem 1 admits a unique solution uVu\in V.

Proof.

Assume g=0g=0. We have

0=g,uΓ=B(u,u)=𝒯u,uΓ=Mu^,u^.0=\Im\,\langle g,u\rangle_{\Gamma}=-\Im\,B(u,u)=\Im\,\left\langle\mathcal{T}u,u\right\rangle_{\Gamma}=\Im\,\left\langle M\hat{u},\hat{u}\right\rangle_{\mathbb{R}}.

Combing (2.12) and (2.17) gives

Mu^,u^=i(PS)(μξγpω2μξ2ω2μξ2μξγs)(ξγs¯γp¯ξ)(P¯S¯)dξ.\left\langle M\hat{u},\hat{u}\right\rangle_{\mathbb{R}}=\int_{\mathbb{R}}i\left(\begin{array}[]{cc}P\\ S\end{array}\right)^{\top}\left(\begin{array}[]{cc}\mu\xi\gamma_{p}&\omega^{2}-\mu\xi^{2}\\ \omega^{2}-\mu\xi^{2}&-\mu\xi\gamma_{s}\end{array}\right)\left(\begin{array}[]{cc}\xi&\bar{\gamma_{s}}\\ \bar{\gamma_{p}}&-\xi\end{array}\right)\left(\begin{array}[]{cc}\bar{P}\\ \bar{S}\end{array}\right)\,\mathrm{d}\xi.

Denote the matrix by

A=(μξγpω2μξ2ω2μξ2μξγs)(ξγs¯γp¯ξ).A=\left(\begin{array}[]{cc}\mu\xi\gamma_{p}&\omega^{2}-\mu\xi^{2}\\ \omega^{2}-\mu\xi^{2}&-\mu\xi\gamma_{s}\end{array}\right)\left(\begin{array}[]{cc}\xi&\bar{\gamma_{s}}\\ \bar{\gamma_{p}}&-\xi\end{array}\right).

Taking the real part implies

A={0 if |ξ|>ks,ω2diag(0,γs) if kp<|ξ|ks,ω2diag(γp,γs) if |ξ|kp.\Re\,A=\left\{\begin{array}[]{ccc}0&\text{ if }|\xi|>k_{s},\\ \omega^{2}\text{diag}(0,\gamma_{s})&\qquad\text{ if }\,k_{p}<|\xi|\leq k_{s},\\ \omega^{2}\text{diag}(\gamma_{p},\gamma_{s})&\text{ if }\,|\xi|\leq k_{p}.\end{array}\right.

So we have

𝒯u,uΓ=ω2(|ξ|kpγp(ξ)|P(ξ)|2dξ+|ξ|ksγs(ξ)|S(ξ)|2dξ)=0.\Im\langle\mathcal{T}u,u\rangle_{\Gamma}=\omega^{2}\left(\int_{|\xi|\leq k_{p}}\gamma_{p}(\xi)|P(\xi)|^{2}\,\mathrm{d}\xi+\int_{|\xi|\leq k_{s}}\gamma_{s}(\xi)|S(\xi)|^{2}\,\mathrm{d}\xi\right)=0. (3.2)

Hence

P(ξ)=S(ξ)=0for|ξ|kp,P(\xi)=S(\xi)=0\quad\text{for}\quad|\xi|\leq k_{p},

which implies

u^(ξ,0)=0for|ξ|kp.\hat{u}(\xi,0)=0\quad\text{for}\quad|\xi|\leq k_{p}.

Since uu has compact support on {y=0}\{y=0\}, u^(ξ,0)\hat{u}(\xi,0) is analytic with respect to ξ\xi. By unique continuation, u^(ξ,0)=0\hat{u}(\xi,0)=0 for ξ\xi\,\in\mathbb{R}. So u(x,0)=0u(x,0)=0 and Tu=𝒯u=0Tu=\mathcal{T}u=0 on Γ\Gamma. By Holmgren’s uniqueness theorem, u=0u=0 in DD which implies uniqueness. Finally, combing Lemma 3.3 and Fredholm alternative yields existence. ∎

Remark. Theorem 3.1 shows that there admits a unique solution uu to Variation problem 1 in bounded domain DD. We can extend us=uuiuru_{s}=u-u_{i}-u_{r} to {x2>0}\{x_{2}>0\} by (2.14), though it may not satisfy Kupradze-Sommerfeld radiation condition. In fact, the condition that usu_{s} satisfies (2.14) is weaker than Kupradze-Sommerfeld radiation condition of half-plane (2.9)-(2.10) (see [15]). But we can extend usu_{s} by

us=ΓTzGD(x,y)us(y)ds(y),x{x2>0},u_{s}=\int_{\Gamma}T_{z}G_{D}(x,y)u_{s}(y)\,\text{d}\,s(y),\quad x\in\{x_{2}>0\}, (3.3)

where GD(x,y)G_{D}(x,y) is the half-space Green tensor for Dirichlet problem. According to the property of Green tensor we can verify that the corresponding uu is the solution to Drichlet problem in D{x20}D\cup\{x_{2}\geq 0\} (see [15]). Then by representation theorem we know that any solution to Dirichlet problem in D{x20}D\cup\{x_{2}\geq 0\} satisfies (3.3), which means the extension is unique.

4 Existence and uniqueness for Neumann problem

Recalling the two variation problems in Section 2, we can find that the main difference between Dirichlet problem and Neumann one is that MM in (2.17) is replaced by M1M^{-1}, i.e.,

M1=ρ(ξ)id(ξ)(ω2γsξω2ξμρξω2+ξμρω2γp),M^{-1}=\frac{\rho(\xi)}{id(\xi)}\left(\begin{array}[]{cc}\omega^{2}\gamma_{s}&\xi\omega^{2}-\xi\mu\rho\\ -\xi\omega^{2}+\xi\mu\rho&\omega^{2}\gamma_{p}\end{array}\right), (4.1)

where

ρ=ξ2+γsγp,d=ω4γsγp+(ξω2ξμρ)2.\rho=\xi^{2}+\gamma_{s}\gamma_{p},\,d=\omega^{4}\gamma_{s}\gamma_{p}+(\xi\omega^{2}-\xi\mu\rho)^{2}.

This section shows that M1M^{-1} has similar properties to MM though it is more complicated. Before proceed to the uniqueness, we give the following lemma to show that Λ\varLambda is continuous, which implies the sesquilinear form is also continuous.

Lemma 4.1.

The NtD operator Λ\varLambda is a bounded linear operator from H01/2(Γ)2H^{-1/2}_{0}(\Gamma)^{2} to H1/2(Γ)2H^{1/2}(\Gamma)^{2}.

Proof.
Λ2=supuW,u0ΛuH1/2(Γ)22uH1/2(Γ)22=supuW,u0|M1u^|2(1+ξ2)1/2|u^|2(1+ξ2)1/2.\|\varLambda\|^{2}=\sup\limits_{u\in W,\,u\not=0}\frac{\|\varLambda u\|^{2}_{H^{1/2}(\Gamma)^{2}}}{\|u\|^{2}_{H^{-1/2}(\Gamma)^{2}}}=\sup\limits_{u\in W,\,u\not=0}\frac{\int_{\mathbb{R}}|M^{-1}\hat{u}|^{2}(1+\xi^{2})^{1/2}}{\int_{\mathbb{R}}|\hat{u}|^{2}(1+\xi^{2})^{-1/2}}. (4.2)

When |ξ|+|\xi|\to+\infty, it is easy to calculate

γpi|ξ|,γsi|ξ|,and(ξ2+γsγp)kp2+ks22.\gamma_{p}\sim i|\xi|,\,\gamma_{s}\sim i|\xi|,\,\text{and}\,(\xi^{2}+\gamma_{s}\gamma_{p})\sim\frac{k^{2}_{p}+k^{2}_{s}}{2}.

It turns out that

[ξω2ξμ(ξ2+γsγp)]ξ(ω2ks2+kp22),\left[\xi\omega^{2}-\xi\mu(\xi^{2}+\gamma_{s}\gamma_{p})\right]\sim\xi\left(\omega^{2}-\frac{k_{s}^{2}+k_{p}^{2}}{2}\right), (4.3)

and

[ω4γsγp+(ξω2ξμ(ξ2+γsγp))2]ξ2[ω4(ω2μks2+kp22)2].\left[\omega^{4}\gamma_{s}\gamma_{p}+(\xi\omega^{2}-\xi\mu(\xi^{2}+\gamma_{s}\gamma_{p}))^{2}\right]\sim\xi^{2}\left[\omega^{4}-\left(\omega^{2}-\mu\frac{k^{2}_{s}+k^{2}_{p}}{2}\right)^{2}\right]. (4.4)

Combing (4.1) and (4.3)-(4.4) gives estimate of M1\|M^{-1}\|,

M1C(1+ξ2)1.\|M^{-1}\|\leq C(1+\xi^{2})^{-1}.

Inserting it into (4.2) yields ΛC\|\varLambda\|\leq C, where CC depends on ω,μ\omega,\,\mu and λ\lambda. ∎

Then we consider M1-\Re\,M^{-1} similarly as Section 2.

Lemma 4.2.

M1(ξ)>0-\Re\,M^{-1}(\xi)>0 for sufficiently large |ξ||\xi|.

Proof.

Taking real part of M1M^{-1} for |ξ|>ks|\xi|>k_{s}, we have

M1=ρ(ξ)d(ξ)(ω2|γs|i(ξω2ξμρ)i(ξω2ξμρ)ω2|γp|),\Re\,M^{-1}=\frac{\rho(\xi)}{d(\xi)}\left(\begin{array}[]{cc}\omega^{2}|\gamma_{s}|&-i(\xi\omega^{2}-\xi\mu\rho)\\ i(\xi\omega^{2}-\xi\mu\rho)&\omega^{2}|\gamma_{p}|\end{array}\right),

with

ρ(ξ)=ξ2|γs||γp|>0,\rho(\xi)=\xi^{2}-|\gamma_{s}||\gamma_{p}|>0,

and

d(ξ)=ω4|γs||γp|+(ξω2ξμρ)2.d(\xi)=-\omega^{4}|\gamma_{s}||\gamma_{p}|+(\xi\omega^{2}-\xi\mu\rho)^{2}.

In order to prove M1(ξ)>0-\Re M^{-1}(\xi)>0, it suffices to verify ρ(ξ)d(ξ)ω2|γs|<0\frac{\rho(\xi)}{d(\xi)}\omega^{2}|\gamma_{s}|<0 and detM1>0\det{-\Re M^{-1}}>0. Recall when |ξ||\xi|\to\infty, γsi|ξ|,γpi|ξ|\gamma_{s}\sim i|\xi|,\,\gamma_{p}\sim i|\xi| and ρks2+kp22\rho\sim\frac{k^{2}_{s}+k^{2}_{p}}{2}. It turns out that

dξ2((ω2μks2+kp22)2ω4)=ω2(μ4μ+2λ32)<0,d\sim\xi^{2}\left(\left(\omega^{2}-\mu\frac{k^{2}_{s}+k^{2}_{p}}{2}\right)^{2}-\omega^{4}\right)=\omega^{2}\left(\frac{\mu}{4\mu+2\lambda}-\frac{3}{2}\right)<0,

which means ρ(ξ)d(ξ)ω2|γs|<0\frac{\rho(\xi)}{d(\xi)}\omega^{2}|\gamma_{s}|<0 for sufficiently large |ξ||\xi|. Direct calculation gives

det(M1)=ρ2d2(ω4|γs||γp|(ξω2ξμρ)2)>0,\det(-\Re\,M^{-1})=\frac{\rho^{2}}{d^{2}}(\omega^{4}|\gamma_{s}||\gamma_{p}|-(\xi\omega^{2}-\xi\mu\rho)^{2})>0,

for sufficiently large |ξ||\xi|. So M1-\Re\,M^{-1} is positive definite for sufficiently large |ξ||\xi|. ∎

Using the above lemma, we can estimate (u,u)\Re\,\mathcal{B}(u,u) and then get the following inequality.

Lemma 4.3.

For any uWu\in W, we have (u,u)>C1uL2(D)22C2uL2(D)22\Re\,\mathcal{B}(u,u)>C_{1}\|\nabla u\|^{2}_{L^{2}(D)^{2}}-C_{2}\|u\|^{2}_{L^{2}(D)^{2}}, where C1,C2C_{1},\,C_{2} are positive constant and they both depend on λ,μ\lambda,\,\mu and ω\omega.

Proof.

Similar to Section 3, for uWu\in W, we have

(u,u)μuL2(D)22ω2uL2(D)22Tu,Λ(Tu)Γ.\Re\,\mathcal{B}(u,u)\geq\mu\|\nabla u\|^{2}_{L^{2}(D)^{2}}-\omega^{2}\|u\|^{2}_{L^{2}(D)^{2}}-\Re\,\langle Tu,\varLambda(Tu)\rangle_{\Gamma}. (4.5)

By Lemma 4.2, there exists k0>0k_{0}>0 such that M1(ξ)>0-\Re\,M^{-1}(\xi)>0 for |ξ|>k0|\xi|>k_{0}. So it can be deduced that

Tu,Λ(Tu)Γ|ξ|k0Tu^M1Tu^¯dξC|ξ|k0M1|Tu^|2dξ.-\Re\,\langle Tu,\varLambda(Tu)\rangle_{\Gamma}\geq-\Re\,\int_{|\xi|\leq k_{0}}\widehat{Tu}\cdot\overline{M^{-1}\widehat{Tu}}\,\mathrm{d}\,\xi\geq-C\int_{|\xi|\leq k_{0}}\|M^{-1}\||\widehat{Tu}|^{2}\,\mathrm{d}\,\xi. (4.6)

Noting for |ξ|k0|\xi|\leq k_{0},

M1C(1+ξ2)1/2C(1+ξ2)1.\|M^{-1}\|\leq C(1+\xi^{2})^{-1/2}\leq C(1+\xi^{2})^{-1}.

So we have

|ξ|k0M1|Tu^|2dξC|ξ|k0(1+ξ2)1|Tu^|2dξCTuH1(Γ).2\int_{|\xi|\leq k_{0}}\|M^{-1}\||\widehat{Tu}|^{2}\,\mathrm{d}\,\xi\leq C\int_{|\xi|\leq k_{0}}(1+\xi^{2})^{-1}|\widehat{Tu}|^{2}\,\mathrm{d}\,\xi\leq C\|Tu\|^{2}_{H^{-1}(\Gamma).} (4.7)

By the trace theorem and the interpolation inequality, we obtain

TuH1(Γ)22CuL2(Γ)22CuH1/2(D)22εuH1(D)22+C(ε)uL2(D)22.\|Tu\|^{2}_{H^{-1}(\Gamma)^{2}}\leq C\|u\|^{2}_{L^{2}(\Gamma)^{2}}\leq C\|u\|^{2}_{H^{1/2}(D)^{2}}\leq\varepsilon\|u\|^{2}_{H^{1}(D)^{2}}+C(\varepsilon)\|u\|^{2}_{L^{2}(D)^{2}}. (4.8)

Here C>0C>0 are different constants depending on μ,λ\mu,\,\lambda and ω\omega, ε>0\varepsilon>0 is sufficiently small and C(ε)>0C(\varepsilon)>0 is dependent on μ,λ,ω\mu,\,\lambda,\,\omega and ε\varepsilon. By combining (4.5)-(4.8), we obtain the inequality

(u,u)μuL2(D)22ω2uL2(D)22CεuH1(D)22C(ε)uL2(D)22.\Re\,\mathcal{B}(u,u)\geq\mu\|\nabla u\|^{2}_{L^{2}(D)^{2}}-\omega^{2}\|u\|^{2}_{L^{2}(D)^{2}}-C\varepsilon\|u\|^{2}_{H^{1}(D)^{2}}-C(\varepsilon)\|u\|^{2}_{L^{2}(D)^{2}}.

As ε\varepsilon is sufficiently small, it turns out that

(u,u)C1uL2(D)22C2uL2(D)22.\Re\,\mathcal{B}(u,u)\geq C_{1}\|\nabla u\|^{2}_{L^{2}(D)^{2}}-C_{2}\|u\|^{2}_{L^{2}(D)^{2}}.

Next, we can proceed to the uniqueness for the Neumann problem. Though M1M^{-1} is more complicated than MM, it is difficult to directly give the representation of the imaginary part of the sesquilinear form like (3.2). But the following proof shows that it is not necessary to get exact representation of the imaginary part of the sesquilinear form.

Theorem 4.1.

Variation problem 2 admits a unique solution uWu\in W.

Proof.

Similar to Section 3, we will prove the uniqueness, which yields existence. Assume h=0h=0 (i.e., u=usu=u_{s}) which implies g=0g=0. Taking imaginary part of variation formula gives

(u,u)=Tu,Λ(Tu)Γ=Tu^,M1Tu^Γ=M1Tu^,Tu^Γ=0.\Im\,\mathcal{B}(u,u)=-\Im\,\langle Tu,\varLambda(Tu)\rangle_{\Gamma}=-\Im\,\langle\widehat{Tu},M^{-1}\widehat{Tu}\rangle_{\Gamma}=\langle\Im\,M^{-1}\widehat{Tu},\widehat{Tu}\rangle_{\Gamma}=0.

For |ξ|>ks|\xi|>k_{s}, we have

ρ=ξ2|γs||γp|,d=ω4|γs||γp|+(ξω2ξμρ)2.\rho=\xi^{2}-|\gamma_{s}||\gamma_{p}|,\,d=-\omega^{4}|\gamma_{s}||\gamma_{p}|+(\xi\omega^{2}-\xi\mu\rho)^{2}.

Then take imaginary part of (4.1) to get M1=0\Im M^{-1}=0. For |ξ|kp|\xi|\leq k_{p}, we have

ρ=ξ2+|γs|γp|>0,d=ω4|γs||γp|+(ξω2ξμρ)2>0.\rho=\xi^{2}+|\gamma_{s}|\gamma_{p}|>0,\,d=\omega^{4}|\gamma_{s}||\gamma_{p}|+(\xi\omega^{2}-\xi\mu\rho)^{2}>0.

Simple calculation gives

M1=ρ(ξ)d(ξ)(ω2|γs|00ω2|γp|).\Im M^{-1}=-\frac{\rho(\xi)}{d(\xi)}\left(\begin{array}[]{cc}\omega^{2}|\gamma_{s}|&0\\ 0&\omega^{2}|\gamma_{p}|\end{array}\right).

It turns out that M1<0\Im M^{-1}<0. For kp<|ξ|ksk_{p}<|\xi|\leq k_{s}, we have

ρ=ξ2+i|γs|γp|,d=ω4|γs||γp|i+(ξω2ξμρ)2.\rho=\xi^{2}+i|\gamma_{s}|\gamma_{p}|,\,d=\omega^{4}|\gamma_{s}||\gamma_{p}|i+(\xi\omega^{2}-\xi\mu\rho)^{2}.

Direct calculation gives

M1=(a1b1+b2b1b2a2),\Im M^{-1}=\left(\begin{array}[]{cc}a_{1}&b_{1}+b_{2}\\ -b_{1}-b_{2}&a_{2}\end{array}\right),

where

a1=ω6|γs|3|γp|2+2ω2|γp|2|γs|3ξ2μζξ4ω2|γs|ζ2+ω2ξ4μ2|γp|2|γs|3,a_{1}=-\omega^{6}|\gamma_{s}|^{3}|\gamma_{p}|^{2}+2\omega^{2}|\gamma_{p}|^{2}|\gamma_{s}|^{3}\xi^{2}\mu\zeta-\xi^{4}\omega^{2}|\gamma_{s}|\zeta^{2}+\omega^{2}\xi^{4}\mu^{2}|\gamma_{p}|^{2}|\gamma_{s}|^{3},
a2=ξω6|γp|2|γs|+2ξ4ω2μζ|γs||γp|2+ω2|γp|2|γs|ξ2ζ2ξ2μ2|γp|4|γs|3,a_{2}=-\xi\omega^{6}|\gamma_{p}|^{2}|\gamma_{s}|+2\xi^{4}\omega^{2}\mu\zeta|\gamma_{s}||\gamma_{p}|^{2}+\omega^{2}|\gamma_{p}|^{2}|\gamma_{s}|\xi^{2}\zeta^{2}-\xi^{2}\mu^{2}|\gamma_{p}|^{4}|\gamma_{s}|^{3},
b1=iξ3ω4ζ|γs||γp|2iξ5μζ2|γs||γp|iξ3ζ3|γs||γp|+iξ3μ2ζ|γs|3|γp|3,b_{1}=i\xi^{3}\omega^{4}\zeta|\gamma_{s}||\gamma_{p}|-2i\xi^{5}\mu\zeta^{2}|\gamma_{s}||\gamma_{p}|-i\xi^{3}\zeta^{3}|\gamma_{s}||\gamma_{p}|+i\xi^{3}\mu^{2}\zeta|\gamma_{s}|^{3}|\gamma_{p}|^{3},
b2=iω4ξμ|γp|3|γs|32iξ3μ2|γp|3|γs|3ζ+iξ5μζ2|γs||γp|iξ5μ3|γp|3|γs|3,b_{2}=i\omega^{4}\xi\mu|\gamma_{p}|^{3}|\gamma_{s}|^{3}-2i\xi^{3}\mu^{2}|\gamma_{p}|^{3}|\gamma_{s}|^{3}\zeta+i\xi^{5}\mu\zeta^{2}|\gamma_{s}||\gamma_{p}|-i\xi^{5}\mu^{3}|\gamma_{p}|^{3}|\gamma_{s}|^{3},

with

ζ=ω2ξ2μ.\zeta=\omega^{2}-\xi^{2}\mu.

In order to prove M10\Im M^{-1}\leq 0, it suffices to verify a1<0a_{1}<0 and detM1=0\det{\Im M^{-1}}=0. By direct calculation,

a1=(|γs|2|γp|2ζ2ξ4ζ2)ω2|γs|<0,a_{1}=(-|\gamma_{s}|^{2}|\gamma_{p}|^{2}\zeta^{2}-\xi^{4}\zeta^{2})\omega^{2}|\gamma_{s}|<0,

and

detM1=a1a2+(b1+b2)2=d1|γs|6|γp|6+d2|γs|4|γp|4+d3|γs|2|γp|2,\det{\Im M^{-1}}=a_{1}a_{2}+(b_{1}+b_{2})^{2}=d_{1}|\gamma_{s}|^{6}|\gamma_{p}|^{6}+d_{2}|\gamma_{s}|^{4}|\gamma_{p}|^{4}+d_{3}|\gamma_{s}|^{2}|\gamma_{p}|^{2},

where

d1\displaystyle d_{1} =ω4ξ2μ2(ω4ξ2μ(2ω2ξ2μ))(ζξμω2)2=0,\displaystyle=\omega^{4}\xi^{2}\mu^{2}(\omega^{4}-\xi^{2}\mu(2\omega^{2}-\xi^{2}\mu))-(\zeta\xi\mu\omega^{2})^{2}=0,
d2\displaystyle d_{2} =ξ6ω4μ2(ω4ξ2μ(2ω2ξ2μ))+ξ6ω4μ2ζ22ξ6μ2ω4ζ2=0,\displaystyle=\xi^{6}\omega^{4}\mu^{2}(\omega^{4}-\xi^{2}\mu(2\omega^{2}-\xi^{2}\mu))+\xi^{6}\omega^{4}\mu^{2}\zeta^{2}-2\xi^{6}\mu^{2}\omega^{4}\zeta^{2}=0,
d3\displaystyle d_{3} =ξ10ζ2ω4μ2(ξ3ω4ζξ3ω2ζ2)2=0.\displaystyle=\xi^{10}\zeta^{2}\omega^{4}\mu^{2}-(\xi^{3}\omega^{4}\zeta-\xi^{3}\omega^{2}\zeta^{2})^{2}=0.

Therefore it turns out that detM1=0\det\Im M^{-1}=0, and thus M10\Im M^{-1}\leq 0.

In summary,

M1{=0if|ξ|>ks,0ifkp<|ξ|ks,<0if|ξ|kp.\Im\,M^{-1}\left\{\begin{array}[]{ccc}=0&\text{if}\,|\xi|>k_{s},\\ \leq 0&\text{if}\,k_{p}<|\xi|\leq k_{s},\\ <0&\text{if}\,|\xi|\leq k_{p}.\end{array}\right.

Hence Tu^=0\widehat{Tu}=0 for |ξ|kp|\xi|\leq k_{p}. Similarly to Section 3, by unique continuation and Holmgren’s uniqueness theorem, u=0u=0 in DD which implies uniqueness. Combing Lemma 4.3 and Fredholm alternative yields existence.

Remark. Similarly to Section 2, we can extend usu_{s} to {x2>0}\{x_{2}>0\} by

us=ΓGN(x,y)Tus(y)ds(y)x{x2>0},u_{s}=-\int_{\Gamma}G_{N}(x,y)Tu_{s}(y)\,\text{d}s(y)\quad x\in\{x_{2}>0\}, (4.9)

where GN(x,y)G_{N}(x,y) is the half-space Green tensor for Neumann problem. Obviously by representation theorem, this extension is unqiue. Hence we get the unique solution to Neumann problem in D{x20}D\cup\{x_{2}\geq 0\}.

5 Inverse cavity problem

In this section, an inverse cavity problem, i.e., to reconstruct the unknown cavity SS from u|Γu|_{\Gamma}, is considered by investigating the Fre´\acute{\rm e}chet derivative of the solution operator. For Dirichlet problem, the Freche´\acute{\rm e}t derivative implies a local stability result of inverse cavity problem directly. This can be deduced by the fact that when the Dirichlet data of Freche´\acute{\rm e}t derivative on the boundary of the cavity vanishes, the solution uu to original problem is zero. However, for Neumann case, the Freche´\acute{\rm e}t derivative satisfies different boundary value condition, which results in major difference from the Dirichlet case.

5.1 Dirichlet case

Let SS be C2C^{2} and h(x)C2(S,)2h(x)\in C^{2}(S,\mathbb{R})^{2} satisfying h(x)|SΓ=0h(x)|_{S\cap\Gamma}=0. Denote by Sh={x+h(x):xS}S_{h}=\{x+h(x):x\in S\} and DhD_{h} the domain with boundary SΓS\cup\Gamma. Similarly denote Vh={uH1(Dh)2:u=0,onSh}V_{h}=\{u\in H^{1}(D_{h})^{2}:u=0,\,\text{on}\,S_{h}\}. Define the solution operator 𝒰\mathcal{U} : C2(S,2)H1/2(Γ)2C^{2}(S,\mathbb{R}^{2})\to H^{1/2}(\Gamma)^{2} by

𝒰(h)=uh|Γ,\mathcal{U}(h)=u_{h}|_{\Gamma},

where uhu_{h} is the solution to variation problem for Dirichlet boundary condition, i.e.,

Bh(uh,vh)=(g,vh)Γ,vhVhB_{h}(u_{h},v_{h})=(g,v_{h})_{\Gamma},\quad\forall v_{h}\in V_{h} (5.1)

with

Bh(uh,vh)=Dh(uh,v¯h)ω2uhv¯dxΓ𝒯uv¯ds.B_{h}(u_{h},v_{h})=\int_{D_{h}}\mathcal{E}(u_{h},\bar{v}_{h})-\omega^{2}u_{h}\cdot\bar{v}\,\mathrm{d}x-\int_{\Gamma}\mathcal{T}u\cdot\bar{v}\,\mathrm{d}s. (5.2)

Extend h(x)h(x) to C2(D¯,2)C^{2}(\bar{D},\mathbb{R}^{2}) and still denote it by hh such that h(x)|Γ=0h(x)|_{\Gamma}=0, which gives

hH1,(D)2ChH1,(S)2.\|h\|_{H^{1,\infty}(D)^{2}}\leq C\|h\|_{H^{1,\infty}(S)^{2}}. (5.3)

Denote h\mathcal{H}_{h} by h(y)=y+h(y),yD\mathcal{H}_{h}(y)=y+h(y),\quad y\in D. Obviously, it is invertible for sufficiently small h1,\|h\|_{1,\infty}. Then taking the variable transform x=h(y)x=\mathcal{H}_{h}(y) implies

Bh(uh,vh)=\displaystyle B_{h}(u_{h},v_{h})= μDj=12u~hj𝒥h1𝒥h1v~¯hjdet𝒥hdy\displaystyle\mu\int_{D}\sum_{j=1}^{2}\nabla\tilde{u}_{h_{j}}\mathcal{J}_{\mathcal{H}_{h}^{-1}}\mathcal{J}_{\mathcal{H}_{h}^{-1}}^{\top}\nabla\bar{\tilde{v}}_{h_{j}}\det{\mathcal{J}_{\mathcal{H}_{h}}}\,\text{d}y
+(λ+μ)D(u~h:𝒥h1)(v~¯h:𝒥h1)det𝒥hdy\displaystyle+(\lambda+\mu)\int_{D}(\nabla\tilde{u}_{h}:\mathcal{J}_{\mathcal{H}_{h}^{-1}})(\nabla\bar{\tilde{v}}_{h}:\mathcal{J}_{\mathcal{H}_{h}^{-1}}^{\top})\det{\mathcal{J}_{\mathcal{H}_{h}}}\,\text{d}y
ω2Du~hv~¯hdet𝒥hdyΓ𝒯u~hv~¯hds(y):=Bh(u~h,v~h),\displaystyle-\omega^{2}\int_{D}\tilde{u}_{h}\cdot\bar{\tilde{v}}_{h}\det{\mathcal{J}_{\mathcal{H}_{h}}}\,\text{d}y-\int_{\Gamma}\mathcal{T}\tilde{u}_{h}\cdot\bar{\tilde{v}}_{h}\,\text{d}s(y):=B^{h}(\tilde{u}_{h},\tilde{v}_{h}),

where u~h=uhhV\tilde{u}_{h}=u_{h}\circ\mathcal{H}_{h}\in V, v~h=vhhV\tilde{v}_{h}=v_{h}\circ\mathcal{H}_{h}\in V. Hence BhB^{h} is a sesquilinear form on V×VV\times V. Thus the variation formula (5.1) becomes

Bh(u~h,v)=(g,v)Γ,vV.B^{h}(\tilde{u}_{h},v)=(g,v)_{\Gamma},\quad\forall v\in V. (5.4)

Next we proceed to derive the Fre´\acute{\rm e}chet derivative of the solution operator 𝒰\mathcal{U}. Let 𝒰(0)\mathcal{U}^{\prime}(0) : C2(S,)2H1/2(Γ)2C^{2}(S,\mathbb{R})^{2}\to H^{1/2}(\Gamma)^{2} is the Fre´\acute{\rm e}chet derivative of 𝒰\mathcal{U} at 0. Notice that C2(S,)2C^{2}(S,\mathbb{R})^{2} is equipped with norm H1,(S)2\|\cdot\|_{H^{1,\infty}(S)^{2}}. The following theorem shows for hC2(S,)2h\in C^{2}(S,\mathbb{R})^{2}, 𝒰(0)h\mathcal{U}^{\prime}(0)h satisfies the homogeneous Navier equation in DD, the homogeneous TBC on Γ\Gamma and an inhomogeneous Dirichlet condition (5.7) on SS.

Theorem 5.1.

Suppose hC2(S,)2h\in C^{2}(S,\mathbb{R})^{2} satisfying h=0h=0 on SΓS\cap\Gamma. Let uu^{*} is the solution to the problem

(Δ+ω2)u=0,\displaystyle(\Delta^{*}+\omega^{2})u^{*}=0,\quad inD,\displaystyle{\rm in}\quad D, (5.5)
Tu=𝒯u,\displaystyle Tu^{*}=\mathcal{T}u^{*},\quad onΓ,\displaystyle{\rm on}\quad\Gamma, (5.6)
u=(nh)nu,\displaystyle u^{*}=-(n\cdot h)\partial_{n}u,\quad onS.\displaystyle{\rm on}\quad S. (5.7)

Then 𝒰(0)h=u|Γ\mathcal{U}^{\prime}(0)h=u^{*}|_{\Gamma}.

Proof.

Let wh=huw_{h}=h\cdot\nabla u. Then h|Γ=0h|_{\Gamma}=0 implies wh|Γ=0w_{h}|_{\Gamma}=0. So wh+uVw_{h}+u^{*}\in V and (wh+u)|Γ=u|Γ(w_{h}+u^{*})|_{\Gamma}=u^{*}|_{\Gamma}. By trace theorem,

uhuuH1/2(Γ)2Cu~huuwhH1(D)2\|u_{h}-u-u^{*}\|_{H^{1/2}(\Gamma)^{2}}\leq C\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(D)^{2}} (5.8)

with the constant C>0C>0. For convenience, we denote h1,=hH1,(D)2\|h\|_{1,\infty}=\|h\|_{H^{1,\infty}(D)^{2}}. Then combining (5.3) and (5.8) yields we only need to prove

limh1,0u~huuwhH1(D)2h1,=0.\lim_{\|h\|_{1,\infty}\to 0}\frac{\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(D)^{2}}}{\|h\|_{1,\infty}}=0.

For any vVv\in V,

B(u~huuwh,ϕ)=B(u~hu,v)B(u+u,v).B(\tilde{u}_{h}-u-u^{*}-w_{h},\phi)=B(\tilde{u}_{h}-u,v)-B(u^{*}+u,v).

Firstly consider

B(u~hu,v)=B(u~h,v)(g,v)Γ=B(u~h,v)Bh(u~h,v)=i=13Bi(u~h,v),B(\tilde{u}_{h}-u,v)=B(\tilde{u}_{h},v)-(g,v)_{\Gamma}=B(\tilde{u}_{h},v)-B^{h}(\tilde{u}_{h},v)=\sum_{i=1}^{3}B_{i}(\tilde{u}_{h},v),

where

B1(u~h,v)\displaystyle B_{1}(\tilde{u}_{h},v) =μDj=12u~h,j(I2𝒥h1𝒥h1det𝒥h)v¯jdx,\displaystyle=\mu\int_{D}\sum_{j=1}^{2}\nabla\tilde{u}_{h,j}(I_{2}-\mathcal{J}_{\mathcal{H}^{-1}_{h}}\mathcal{J}_{\mathcal{H}^{-1}_{h}}^{\top}\det{\mathcal{J}_{\mathcal{H}_{h}}})\nabla\bar{v}_{j}\,\text{d}x, (5.9)
B2(u~h,v)\displaystyle B_{2}(\tilde{u}_{h},v) =(λ+μ)D(u~h)(v¯)(u~h:𝒥h1)(v¯:𝒥h1)det𝒥hdx,\displaystyle=(\lambda+\mu)\int_{D}(\nabla\cdot\tilde{u}_{h})(\nabla\cdot\bar{v})-(\nabla\tilde{u}_{h}:\mathcal{J}_{\mathcal{H}^{-1}_{h}})(\nabla\bar{{v}}:\mathcal{J}_{\mathcal{H}^{-1}_{h}}^{\top})\det{\mathcal{J}_{\mathcal{H}_{h}}}\,\text{d}x, (5.10)
B3(u~h,v)\displaystyle B_{3}(\tilde{u}_{h},v) =ω2Du~hv¯(det𝒥h1)dx.\displaystyle=\omega^{2}\int_{D}\tilde{u}_{h}\cdot\bar{v}(\det{\mathcal{J}_{\mathcal{H}_{h}}}-1)\,\text{d}x. (5.11)

Consider the Jacobi matrix

𝒥h=I2+h.\mathcal{J}_{\mathcal{H}_{h}}=I_{2}+\nabla h.

Direct calculation implies

det𝒥h=1+h+O(h1,2vH1(D)2)\det{\mathcal{J}_{\mathcal{H}_{h}}}=1+\nabla\cdot h+O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(D)^{2}}) (5.12)

and

𝒥h1=I2h+O(h1,2vH1(D)2).\mathcal{J}_{\mathcal{H}^{-1}_{h}}=I_{2}-\nabla h+O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(D)^{2}}). (5.13)

Combining (5.12) and (5.13) gives

𝒥h1𝒥h1det𝒥h=I2(h+h)(h)I2+O(h1,2vH1(D)2).\mathcal{J}^{\top}_{\mathcal{H}^{-1}_{h}}\mathcal{J}_{\mathcal{H}^{-1}_{h}}\det{\mathcal{J}_{\mathcal{H}_{h}}}=I_{2}-(\nabla h+\nabla h^{\top})-(\nabla\cdot h)I_{2}+O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(D)^{2}}). (5.14)

Insert (5.12)-(5.14) into (5.9)-(5.11),

i=13Bi(u~h,v)=i=13gi(h)(u~h,v)+O(h1,2vH1(D)2),\displaystyle\sum_{i=1}^{3}B_{i}(\tilde{u}_{h},v)=\sum_{i=1}^{3}g_{i}(h)(\tilde{u}_{h},v)+O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(D)^{2}}), (5.15)

where

g1(h)(u~h,v)\displaystyle g_{1}(h)(\tilde{u}_{h},v) =μDj=12u~h,j(h+h(h)I2)v¯jdx,\displaystyle=\mu\int_{D}\sum_{j=1}^{2}\nabla\tilde{u}_{h,j}(\nabla h+\nabla h^{\top}-(\nabla\cdot h)I_{2})\nabla\bar{v}_{j}\,\mathrm{d}x, (5.16)
g2(h)(u~h,v)\displaystyle g_{2}(h)(\tilde{u}_{h},v) =(λ+μ)D(u~h)(v¯:h)+(v¯)(u~h:h)\displaystyle=(\lambda+\mu)\int_{D}(\nabla\cdot\tilde{u}_{h})(\nabla\bar{v}:\nabla h^{\top})+(\nabla\cdot\bar{v})(\nabla\tilde{u}_{h}:\nabla h^{\top})
(h)(v¯)(u~h)dx,\displaystyle-(\nabla\cdot h)(\nabla\cdot\bar{v})(\nabla\cdot\tilde{u}_{h})\,\mathrm{d}x, (5.17)
g3(h)(u~h,v)\displaystyle g_{3}(h)(\tilde{u}_{h},v) =ω2D(u~hv¯)(h)dx.\displaystyle=\omega^{2}\int_{D}(\tilde{u}_{h}\cdot\bar{v})(\nabla\cdot h)\,\mathrm{d}x. (5.18)

It is easy to verify

|gi(h)(u~h,v)gi(h)(u,v)|=O(h1,vH1(D)2u~huH1(D)2).|g_{i}(h)(\tilde{u}_{h},v)-g_{i}(h)(u,v)|=O(\|h\|_{1,\infty}\|v\|_{H^{1}(D)^{2}}\|\tilde{u}_{h}-u\|_{H^{1}(D)^{2}}). (5.19)

Insert (5.19) into (5.15)-(5.18),

B(u~hu,v)\displaystyle B(\tilde{u}_{h}-u,v) =g1(h)(u,v)+g2(h)(u,v)+g3(h)(u,v)+\displaystyle=g_{1}(h)(u,v)+g_{2}(h)(u,v)+g_{3}(h)(u,v)+
O(h1,2vH1(D)2)+O(h1,vH1(D)2u~huH1(D)2)\displaystyle O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(D)^{2}})+O(\|h\|_{1,\infty}\|v\|_{H^{1}(D)^{2}}\|\tilde{u}_{h}-u\|_{H^{1}(D)^{2}}) (5.20)

For g1(h)(u,v)g_{1}(h)(u,v), applying the identity

u(h+h(h)I2)v=\displaystyle\nabla u(\nabla h+\nabla h^{\top}-(\nabla\cdot h)I_{2})\nabla v= {h(u)v+(hv)u(uv)h}\displaystyle\nabla\cdot\{h\cdot(\nabla u)\nabla v+(h\cdot\nabla v)\nabla u-(\nabla u\cdot\nabla v)h\}
(hv)Δu(hu)Δv\displaystyle-(h\cdot\nabla v)\Delta u-(h\cdot\nabla u)\Delta v

and divergence theorem gives

g1(h)(u,v)\displaystyle g_{1}(h)(u,v) =μj=12{D(huj)Δv¯j+(hv¯j)Δujdx\displaystyle=-\mu\sum_{j=1}^{2}\Big{\{}\int_{D}(h\cdot\nabla u_{j})\Delta\bar{v}_{j}+(h\cdot\nabla\bar{v}_{j})\Delta{u}_{j}\,\mathrm{d}x
S(huj)(nv¯j)+(hv¯j)(nuj)(hn)(v¯juj)ds}.\displaystyle-\int_{S}(h\cdot\nabla u_{j})(n\cdot\nabla\bar{v}_{j})+(h\cdot\nabla\bar{v}_{j})(n\cdot\nabla{u}_{j})-(h\cdot n)(\nabla\bar{v}_{j}\cdot\nabla u_{j})\,\mathrm{d}s\Big{\}}. (5.21)

Considering (Δ+ω2)u=0(\Delta^{*}+\omega^{2})u=0, integration by parts gives

j=12\displaystyle\sum_{j=1}^{2} μD(hv¯j)Δujdx=(λ+μ)D(u)(hv¯)dx\displaystyle\mu\int_{D}(h\cdot\nabla\bar{v}_{j})\Delta{u}_{j}\,\mathrm{d}x=(\lambda+\mu)\int_{D}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})\,\mathrm{d}x
(λ+μ)S(u)(n(hv¯))dsω2D(hv¯)udx.\displaystyle-(\lambda+\mu)\int_{S}(\nabla\cdot u)(n\cdot(h\cdot\nabla\bar{v}))\,\mathrm{d}s-\omega^{2}\int_{D}(h\cdot\nabla\bar{v})\cdot u\,\mathrm{d}x. (5.22)

Integration by parts again gives

j=12μ\displaystyle\sum_{j=1}^{2}\mu D(huj)Δv¯jdx=\displaystyle\int_{D}(h\cdot\nabla u_{j})\Delta\bar{v}_{j}\,\mathrm{d}x=
μD(hu):v¯dx+μS(hu)(nv¯)ds.\displaystyle-\mu\int_{D}\nabla(h\cdot\nabla u):\nabla\bar{v}\,\mathrm{d}x+\mu\int_{S}(h\cdot\nabla u)\cdot(n\nabla\bar{v})\,\mathrm{d}s. (5.23)

Denote n=(n1,n2)n=(n_{1},n_{2})^{\top} and τ=(n2,n1)\tau=(-n_{2},n_{1})^{\top}. Notice v|S=0v|_{S}=0 implies τv|S=0\partial_{\tau}v|_{S}=0. This turns out that

i=12S(hv¯)(nuj)(hn)(ujv¯j)ds=0.\sum_{i=1}^{2}\int_{S}(h\cdot\nabla\bar{v})(n\cdot\nabla u_{j})-(h\cdot n)(\nabla u_{j}\cdot\nabla\bar{v}_{j})\,\mathrm{d}s=0. (5.24)

Applying the divergence theorem again and noticing h|Γ=0h|_{\Gamma}=0, v|S=0v|_{S}=0, we obtain

D(h)(uv¯)+u(hv¯)dx\displaystyle\int_{D}(\nabla\cdot h)(u\cdot\bar{v})+u\cdot(h\cdot\nabla\bar{v})\,\mathrm{d}x =D((uv¯)h)(hu)v¯dx\displaystyle=\int_{D}\nabla\cdot((u\cdot\bar{v})h)-(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x
=D(hu)v¯dx.\displaystyle=-\int_{D}(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x. (5.25)

Inserting (5.1)-(5.1) into (5.16) and (5.18) yields

g1(h)(u,v)\displaystyle g_{1}(h)(u,v) +g3(h)(u,v)=\displaystyle+g_{3}(h)(u,v)=
μD(hu):v¯dx(λ+μ)D(u)(hv¯)dx\displaystyle\mu\int_{D}\nabla(h\cdot\nabla u):\nabla\bar{v}\,\mathrm{d}x-(\lambda+\mu)\int_{D}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})\mathrm{d}x
ω2D(hu)v¯dx+(λ+μ)S(u)(n(hv¯))ds.\displaystyle-\omega^{2}\int_{D}(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x+(\lambda+\mu)\int_{S}(\nabla\cdot u)(n\cdot(h\cdot\nabla\bar{v}))\mathrm{d}s. (5.26)

For g2(h)(u,v)g_{2}(h)(u,v), direct calculation gives

D(u)(v¯:h)dx=D(u)(hv¯)(u)(h(v¯))dx\displaystyle\int_{D}(\nabla\cdot u)(\nabla\bar{v}:\nabla h^{\top})\,\mathrm{d}x=\int_{D}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})-(\nabla\cdot u)(h\cdot(\nabla\cdot\nabla\bar{v}^{\top}))\,\mathrm{d}x (5.27)

and similarly

D(v¯)(u:h)dx=D(v¯)(hu)(v¯)(h(u))dx.\displaystyle\int_{D}(\nabla\cdot\bar{v})(\nabla u:\nabla h^{\top})\,\mathrm{d}x=\int_{D}(\nabla\cdot\bar{v})\nabla\cdot(h\cdot\nabla u)-(\nabla\cdot\bar{v})(h\cdot(\nabla\cdot\nabla u^{\top}))\,\mathrm{d}x. (5.28)

Then applying divergence theorem yields

D(v¯)(h(u))dx+D(u)(h(v¯))dx\displaystyle\int_{D}(\nabla\cdot\bar{v})(h\cdot(\nabla\cdot\nabla u^{\top}))\,\mathrm{d}x+\int_{D}(\nabla\cdot u)(h\cdot(\nabla\cdot\nabla\bar{v}^{\top}))\,\mathrm{d}x
+D(u)(v¯)(h)dx=D(h(u)(v¯))dx\displaystyle+\int_{D}(\nabla\cdot u)(\nabla\cdot\bar{v})(\nabla\cdot h)\,\mathrm{d}x=\int_{D}\nabla\cdot(h(\nabla\cdot u)(\nabla\cdot\bar{v}))\,\mathrm{d}x
=S(hn)(u)(v¯)ds.\displaystyle=\int_{S}(h\cdot n)(\nabla\cdot u)(\nabla\cdot\bar{v})\,\mathrm{d}s. (5.29)

Recalling τv|S=0\partial_{\tau}v|_{S}=0, we have

S(hn)(u)(v¯)ds=S(u)(n(hv¯))ds.\displaystyle\int_{S}(h\cdot n)(\nabla\cdot u)(\nabla\cdot\bar{v})\,\mathrm{d}s=\int_{S}(\nabla\cdot u)(n\cdot(h\cdot\nabla\bar{v}))\mathrm{d}s. (5.30)

Together with (5.27)-(5.30) implies

g2(h)(u,v)\displaystyle g_{2}(h)(u,v) =(λ+μ)D(u)(hv¯)+(hu)(v¯)dx\displaystyle=(\lambda+\mu)\int_{D}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})+\nabla\cdot(h\cdot\nabla u)(\nabla\cdot\bar{v})\,\mathrm{d}x
(λ+μ)S(u)(n(hv¯))ds.\displaystyle-(\lambda+\mu)\int_{S}(\nabla\cdot u)(n\cdot(h\cdot\nabla\bar{v}))\,\mathrm{d}s. (5.31)

Now recalling wh=huw_{h}=h\cdot\nabla u, wh|Γ=0w_{h}|_{\Gamma}=0 and combining the representations of gi(h)(u,v)g_{i}(h)(u,v) in (5.1) and (5.1) implies

i=13gi(h)(u,v)=\displaystyle\sum_{i=1}^{3}g_{i}(h)(u,v)=
μD(hu):v¯dx+(λ+μ)D(hu)v¯dxω2D(hu)v¯dx\displaystyle\mu\int_{D}\nabla(h\cdot\nabla u):\nabla\bar{v}\,\mathrm{d}x+(\lambda+\mu)\int_{D}\nabla\cdot(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x-\omega^{2}\int_{D}(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x
=B(wh,v).\displaystyle=B(w_{h},v). (5.32)

Notice uu^{*} satisfies (5.5)-(5.6), so it satisfies the variation formula B(u,v)=0B(u^{*},v)=0 for any vVv\in V. Hence by (5.1) and (5.1) we can obtain

B(u~huuwh,v)h1,vH1(D)2=O(h1,)+O(u~huH1(D)2).\frac{B(\tilde{u}_{h}-u-u^{*}-w_{h},v)}{\|h\|_{1,\infty}\|v\|_{H^{1}(D)^{2}}}=O(\|h\|_{1,\infty})+O(\|\tilde{u}_{h}-u\|_{H^{1}(D)^{2}}). (5.33)

The Dirichlet boundary condition (5.7) and u,u~hVu,\tilde{u}_{h}\in V deduces (u~huuwh)|S=0(\tilde{u}_{h}-u-u^{*}-w_{h})|_{S}=0 so that u~huuwhV\tilde{u}_{h}-u-u^{*}-w_{h}\in V. The using Theorem 3.1 we know the sesquilinear form BB generates an invertible bounded linear operator which is still denoted by BB. By Banach open map theorem, B1B^{-1} : VVV^{*}\to V is also a bounded linear operator which implies

u~huuwhH1(D)2CsupvV,v0|B(u~huuwh,v)|vH1(D)2.\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(D)^{2}}\leq C\sup_{v\in V,v\neq 0}\frac{|B(\tilde{u}_{h}-u-u^{*}-w_{h},v)|}{\|v\|_{H^{1}(D)^{2}}}. (5.34)

Then combining (5.33)-(5.34) gives

u~huuwhH1(D)2h1,=O(h1,)+O(u~huH1(D)2).\frac{\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(D)^{2}}}{\|h\|_{1,\infty}}=O(\|h\|_{1,\infty})+O(\|\tilde{u}_{h}-u\|_{H^{1}(D)^{2}}).

Hence we only need to prove when h1,0\|h\|_{1,\infty}\to 0, u~huuwhH1(D)20\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(D)^{2}}\to 0. Since Bh(u,v)B(u,v)=B1(u,v)+B2(u,v)+B3(u,v)B^{h}(u,v)-B(u,v)=B_{1}(u,v)+B_{2}(u,v)+B_{3}(u,v). Inserting (5.12)-(5.14) gives

|Bh(u,v)B(u,v)|Ch1,uH1(D)2vH1(D)2.|B^{h}(u,v)-B(u,v)|\leq C\|h\|_{1,\infty}\|u\|_{H^{1}(D)^{2}}\|v\|_{H^{1}(D)^{2}}.

Since

Bh(u~h,v)=(g,v)Γ=B(u,v),B^{h}(\tilde{u}_{h},v)=(g,v)_{\Gamma}=B(u,v),

we obtain

u~huH1(D)2CsupuV,u0supvV,v0|Bh(u,v)B(u,v)|vH1(D)2uH1(D)2Ch1,0.\|\tilde{u}_{h}-u\|_{H^{1}(D)^{2}}\leq C\sup_{u\in V,u\not=0}\sup_{v\in V,v\not=0}\frac{|B^{h}(u,v)-B(u,v)|}{\|v\|_{H^{1}(D)^{2}}\|u\|_{H^{1}(D)^{2}}}\leq C\|h\|_{1,\infty}\to 0.

Hence

limh1,u~huuwhH1(D)2h1,=0,\lim_{\|h\|_{1,\infty}}\frac{\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(D)^{2}}}{\|h\|_{1,\infty}}=0,

which completes the proof. ∎

Next we consider a local stability result for inverse cavity problem. For two domains D1,D2D_{1},\,D_{2} in 2\mathbb{R}^{2}, define the Hausdorff distance between D1D_{1} and D2D_{2} by

dist(D1,D2)=max{ρ(D1,D2),ρ(D2,D1)},dist(D_{1},D_{2})=\max\{\rho(D_{1},D_{2}),\,\rho(D_{2},D_{1})\},

where

ρ(D1,D2)=supxD1infyD2|xy|.\rho(D_{1},D_{2})=\sup_{x\in D_{1}}\inf_{y\in D_{2}}|x-y|.

Take h(x)=kp(x)nh(x)=kp(x)n, where kk\in\mathbb{R} and p(x)C2(S,)p(x)\in C^{2}(S,\mathbb{R}). It is easy to verify dist(D1,D2)=O(h)dist(D_{1},D_{2})=O(h). Our goal is to give the following local stability result.

Theorem 5.2.

Given pC2(S,)p\in C^{2}(S,\mathbb{R}) and k>0k>0 is sufficiently small, then

dist(Dk,D)CukuH1/2(Γ)2,dist(D_{k},D)\leq C\|u_{k}-u\|_{H^{1/2}(\Gamma)^{2}},

where Dk=DkpnD_{k}=D_{kpn}, uk=ukpnu_{k}=u_{kpn} and C>0C>0 is a constant independent with kk.

Proof.

For p(x)0p(x)\equiv 0, it is obvious this conclusion is true. So we can assume p0p\not\equiv 0.

Assume this conclusion is not true. There exists a p(x)C2(S,)p(x)\in C^{2}(S,\mathbb{R}) such that

ukukH1/2(Γ)20,whenk0,\left\|\frac{u_{k^{\prime}}-u}{k^{\prime}}\right\|_{H^{1/2}(\Gamma)^{2}}\to 0,\quad\text{when}\quad k^{\prime}\to 0, (5.35)

where {uk}\{u_{k^{\prime}}\} is some subsequence of {uk}\{u_{k}\}. Applying Theorem 5.1 to h(x)=kp(x)nh(x)=kp(x)n gives

limk0ukuukH1/2(Γ)2=0,\lim_{k\to 0}\left\|\frac{u_{k}-u-u^{*}}{k}\right\|_{H^{1/2}(\Gamma)^{2}}=0, (5.36)

where uu^{*} satisfies

(Δ+ω2)u=0,\displaystyle(\Delta^{*}+\omega^{2})u^{*}=0,\quad inD,\displaystyle{\rm in}\quad D,
Tu=𝒯u,\displaystyle Tu^{*}=\mathcal{T}u^{*},\quad onΓ,\displaystyle{\rm on}\quad\Gamma,
u=hpnu,\displaystyle u^{*}=-hp\partial_{n}u,\quad onS.\displaystyle{\rm on}\quad S.

Let v=u/hv^{*}=u^{*}/h, then vv^{*} satisfies

(Δ+ω2)v=0,\displaystyle(\Delta^{*}+\omega^{2})v^{*}=0,\quad inD,\displaystyle{\rm in}\quad D,
Tv=𝒯v,\displaystyle Tv^{*}=\mathcal{T}v^{*},\quad onΓ,\displaystyle{\rm on}\quad\Gamma,
v=pnu,\displaystyle v^{*}=-p\partial_{n}u,\quad onS.\displaystyle{\rm on}\quad S.

Then by (5.36) we obtain

limh0uhuhvH1/2(Γ)2=0.\lim_{h\to 0}\left\|\frac{u_{h}-u}{h}-v^{*}\right\|_{H^{1/2}(\Gamma)^{2}}=0. (5.37)

Combining (5.35) and (5.37) yields vH1/2(Γ)2=0\|v^{*}\|_{H^{1/2}(\Gamma)^{2}}=0. Hence

Tv=𝒯v=0,onΓ.Tv^{*}=\mathcal{T}v^{*}=0,\quad\text{on}\quad\Gamma.

By unique continuation,

v=0,inD.v^{*}=0,\quad\text{in}\quad D.

Recall

v=pnu,onS,v^{*}=-p\partial_{n}u,\quad{\rm on}\quad S,

p0p\not\equiv 0 and pC2(S,)p\in C^{2}(S,\mathbb{R}), there must exist a continuous part SSS^{\prime}\subset S such that

p0,onSp\not=0,\quad\text{on}\quad S^{\prime}

which implies

nu=0,onS.\partial_{n}u=0,\quad\text{on}\quad S^{\prime}.

Then by unique continuation u=0u=0 in DD which is a contradiction to Tu=𝒯u+gTu=\mathcal{T}u+g on Γ\Gamma. ∎

5.2 Neumann case

Next consider the Neumann problem. Notice that the boundary value problem deduced by the NtD operator has the inhomogeneous boundary condition (2.20), so it is difficult to directly calculate the Fre´\acute{\rm e}chet derivative like Dirichlet problem. Here, we still take the advantage of the TBC given by a DtN operator, though its definition on Γ\Gamma is difficult. Instead, a upper half-circle artificial boundary is taken, see in Figure 2.

Refer to caption
Figure 2: The Neumann case

Take R>0R>0 such that {x2=0,|x|>R}Γc\{x_{2}=0,|x|>R\}\subset\Gamma^{c} in Figure 2. Denote Ω=DΓ{0<x2,|x|<R}\Omega=D\cup\Gamma\{0<x_{2},|x|<R\}, ΓR={x20,|x|=R}\Gamma_{R}=\{x_{2}\geq 0,|x|=R\} and SR=S(Γc{|x|R})S_{R}=S\cup(\Gamma^{c}\cap\{|x|\leq R\}). Let x{|x|>R}x\in\mathbb{R}\cap\{|x|>R\}.

Recalling the scattering wave usu_{s} of Neumann problem satisfies Tus|Γc=0Tu_{s}|_{\Gamma^{c}}=0 and the Green’s representation theorem yields

us=ΓRTyGN(x,y)us(y)GN(x,y)Tyus(y)ds(y),u_{s}=\int_{\Gamma_{R}}T_{y}G_{N}(x,y)\cdot u_{s}(y)-G_{N}(x,y)T_{y}u_{s}(y)\,\mathrm{d}s(y),

where TyT_{y} is the differential operator TT with respect to yy. Taking xΓRx\to\Gamma_{R} gives

12usΓRTyGN(x,y)us(y)+GN(x,y)Tyus(y)=0.\frac{1}{2}u_{s}-\int_{\Gamma_{R}}T_{y}G_{N}(x,y)\cdot u_{s}(y)+G_{N}(x,y)\cdot T_{y}u_{s}(y)=0.

Define the single-layer operator 𝒮\mathcal{S} by

𝒮v(x)=ΓGN(x,y)v(x)ds(y),vH1/2(ΓR)\mathcal{S}v(x)=\int_{\Gamma}G_{N}(x,y)\cdot v(x)\,\mathrm{d}s(y),\quad v\in H^{1/2}(\Gamma_{R})

and double-layer operator 𝒟\mathcal{D} by

𝒟v(x)=ΓRTyGN(x,y)f(y)ds(y),vH1/2(ΓR).\mathcal{D}v(x)=\int_{\Gamma_{R}}T_{y}G_{N}(x,y)f(y)\,\mathrm{d}s(y),\quad v\in H^{-1/2}(\Gamma_{R}).

We can choose RR such that ω2\omega^{2} is not the interior Dirichlet eigenvalue of Δ\Delta^{*} so that 𝒮\mathcal{S} is invertible. Then define the DtN operator

𝒯:=𝒮1(12𝒟),\mathscr{T}:=-\mathcal{S}^{-1}(\frac{1}{2}\mathcal{I}-\mathcal{D}),

where \mathcal{I} is the identity operator. It is easy to verify this DtN operator implies the TBC

Tu=𝒯u+g,onΓRTu=\mathscr{T}u+g,\quad\text{on}\quad\Gamma_{R}

with g=T(ui+ur)(ui+ur)g=T(u_{i}+u_{r})-\mathscr{(}u_{i}+u_{r}).

Then rewrite the boundary value problem as

(Δ+ω2)u=0,\displaystyle(\Delta^{*}+\omega^{2})u=0,\quad inΩ,\displaystyle\text{in}\quad\Omega,
Tu=0,\displaystyle Tu=0,\quad onSR,\displaystyle\text{on}\quad S_{R},
Tu=𝒯u+g,\displaystyle Tu=\mathscr{T}u+g,\quad onΓc.\displaystyle\text{on}\quad\Gamma_{c}.

The variation formula is given by

(u,v)=(g,v)ΓR,vH1(Ω)2,\mathscr{B}(u,v)=(g,v)_{\Gamma_{R}},\quad\forall v\in H^{1}(\Omega)^{2}, (5.38)

where BcB_{c} : H1(Ω)2×H1(Ω)2H^{1}(\Omega)^{2}\times H^{1}(\Omega)^{2}\to\mathbb{C} is defined by

(u,v)=Ω(u,v¯)ω2uv¯dxΓR𝒯uv¯ds.\mathscr{B}(u,v)=\int_{\Omega}\mathcal{E}(u,\bar{v})-\omega^{2}u\cdot\bar{v}\,\mathrm{d}x-\int_{\Gamma_{R}}\mathscr{T}u\cdot\bar{v}\,\mathrm{d}s.

The Dirichlet data on ΓR\Gamma_{R} is directly given from the Neumann data on Γ\Gamma by (4.9). Hence it is enough to consider the reconstruction from the Dirichlet data on ΓR\Gamma_{R}.

Let Ωh=DhΓ{0<x2,|x|<R}\Omega_{h}=D_{h}\cup\Gamma\cup\{0<x_{2},|x|<R\}. Define the solution operator 𝒰N\mathcal{U}_{N} : C2(S,2)H1/2(ΓR)2C^{2}(S,\mathbb{R}^{2})\to H^{1/2}(\Gamma_{R})^{2} by

𝒰N(h)=uh|ΓR,\mathcal{U}_{N}(h)=u_{h}|_{\Gamma_{R}},

where uhu_{h} is solution to

h(uh,vh)=(g,vh)ΓR,vhH1(Ωh)2\mathscr{B}_{h}(u_{h},v_{h})=(g,v_{h})_{\Gamma_{R}},\quad\forall v_{h}\in H^{1}(\Omega_{h})^{2} (5.39)

with

h(uh,vh)=Ωh(uh,v¯h)ω2uhv¯hdxΓR𝒯uhv¯hds.\mathscr{B}_{h}(u_{h},v_{h})=\int_{\Omega_{h}}\mathcal{E}(u_{h},\bar{v}_{h})-\omega^{2}u_{h}\cdot\bar{v}_{h}\,\mathrm{d}x-\int_{\Gamma_{R}}\mathcal{T}u_{h}\cdot\bar{v}_{h}\,\mathrm{d}s. (5.40)

Extend h(x)h(x) to C2(Ω,2)C^{2}(\Omega,\mathbb{R}^{2}) and still denote it by hh such that

h(x)=0,onΓR,h(x)=0,\quad\text{on}\quad\Gamma_{R},

and

hH1,(Ω)2ChH1,(S)2.\|h\|_{H^{1,\infty}(\Omega)^{2}}\leq C\|h\|_{H^{1,\infty}(S)^{2}}.

with constant C>0C>0. Define h\mathcal{H}_{h} by

h(y)=y+h(y),yΩ.\mathcal{H}_{h}(y)=y+h(y),\quad y\in\Omega.

It is invertible for sufficiently small hH1,(Ω)2\|h\|_{H^{1,\infty}(\Omega)^{2}}.

Then taking the variable transform x=h(y)x=\mathcal{H}_{h}(y) in (5.40) implies

h(uh,vh)=h(u~h,v~h).\displaystyle\mathscr{B}_{h}(u_{h},v_{h})=\mathscr{B}^{h}(\tilde{u}_{h},\tilde{v}_{h}).

Here h\mathscr{B}^{h} is the similarly defined as BhB^{h} in Section 5.1 except that DhD_{h}, Γ\Gamma and 𝒯\mathcal{T} are replaced by Ωh\Omega_{h}, ΓR\Gamma_{R} and 𝒯\mathscr{T} respectively. Since u~h\tilde{u}_{h}, v~hH1(Ω)2\tilde{v}_{h}\in H^{1}(\Omega)^{2}, h\mathscr{B}^{h} is a sesquilinear form on H1(Ω)2×H1(Ω)2H^{1}(\Omega)^{2}\times H^{1}(\Omega)^{2}. So the variation formula (5.39) becomes

h(u~h,v)=(g,v)ΓR,vH1(Ωh)2.\mathscr{B}^{h}(\tilde{u}_{h},v)=(g,v)_{\Gamma_{R}},\quad\forall v\in H^{1}(\Omega_{h})^{2}.

Let 𝒰N(0)\mathcal{U}_{N}^{\prime}(0) : C2(S,)2H1/2(ΓR)2C^{2}(S,\mathbb{R})^{2}\to H^{1/2}(\Gamma_{R})^{2} is the Fre´\acute{\rm e}chet derivative of 𝒰N\mathcal{U}_{N} at 0. Then result similar as Theorem 5.1 holds for Neumann case.

Theorem 5.3.

Suppose hC2(S,)2h\in C^{2}(S,\mathbb{R})^{2} satisfying h=0h=0 on SΓS\cap\Gamma. Let uu^{*} is solution to the problem

(Δ+ω2)u=0,\displaystyle(\Delta^{*}+\omega^{2})u^{*}=0,\quad inΩ,\displaystyle{\rm in}\quad\Omega, (5.41)
Tu=𝒯u,\displaystyle Tu^{*}=\mathcal{T}u^{*},\quad onΓR,\displaystyle{\rm on}\quad\Gamma_{R}, (5.42)
Tu=𝒜,\displaystyle Tu^{*}=\mathcal{A},\quad onSR,\displaystyle{\rm on}\quad S_{R}, (5.43)

where 𝒜\mathcal{A} is a distribution defined on H1/2(SR)2H^{1/2}(S_{R})^{2} by

𝒜(v¯)\displaystyle\mathcal{A}(\bar{v}) =ω2SR(uv¯)(hn)ds\displaystyle=\omega^{2}\int_{S_{R}}(u\cdot\bar{v})(h\cdot n)\,\mathrm{d}s
(λ+μ)SR(u)(v¯)(nh)dsμSR(hn)(u:v¯)ds.\displaystyle-(\lambda+\mu)\int_{S_{R}}(\nabla\cdot u)(\nabla\cdot\bar{v})(n\cdot h)\,\mathrm{d}s-\mu\int_{S_{R}}(h\cdot n)(\nabla u:\nabla\bar{v})\,\mathrm{d}s.

Then 𝒰N(0)h=u|Γc\mathcal{U}_{N}^{\prime}(0)h=u^{*}|_{\Gamma_{c}}.

Proof.

Let wh=huw_{h}=h\cdot\nabla u. Then h|ΓR=0h|_{\Gamma_{R}}=0 implies wh|ΓRw_{h}|_{\Gamma_{R}}. So wh+uH1(Ω)2w_{h}+u^{*}\in H^{1}(\Omega)^{2} and (wh+u)|ΓR=u|ΓR(w_{h}+u^{*})|_{\Gamma_{R}}=u^{*}|_{\Gamma_{R}}. Similarly as Theorem 5.1, we only need to prove

limh1,0u~huuwhH1(Ω)2h1,=0,\lim_{\|h\|_{1,\infty}\to 0}\frac{\|\tilde{u}_{h}-u-u^{*}-w_{h}\|_{H^{1}(\Omega)^{2}}}{\|h\|_{1,\infty}}=0,

where 1,=H1(Ω)2\|\cdot\|_{1,\infty}=\|\cdot\|_{H^{1}(\Omega)^{2}}. For any vH1(Ω)2v\in H^{1}(\Omega)^{2},

(u~huuwh,ϕ)=(u~hu,v)(u+wh,v).\mathscr{B}(\tilde{u}_{h}-u-u^{*}-w_{h},\phi)=\mathscr{B}(\tilde{u}_{h}-u,v)-\mathscr{B}(u^{*}+w_{h},v).

Like Section 5.1, we have

(u~hu,v)\displaystyle\mathscr{B}(\tilde{u}_{h}-u,v) =g1(h)(u,v)+g2(h)(u,v)+g3(h)(u,v)+\displaystyle=g_{1}(h)(u,v)+g_{2}(h)(u,v)+g_{3}(h)(u,v)+
O(h1,2vH1(Ω)2)+O(h1,vH1(Ω)2u~huH1(Ω)2),\displaystyle O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(\Omega)^{2}})+O(\|h\|_{1,\infty}\|v\|_{H^{1}(\Omega)^{2}}\|\tilde{u}_{h}-u\|_{H^{1}(\Omega)^{2}}), (5.44)

where gi(h)g_{i}(h) is defined the same as (5.16)-(5.18) except that DD is replaced by Ω\Omega.

For g1(h)(u,v)g_{1}(h)(u,v), similarly as (5.1), we obtain

g1(h)(u,v)\displaystyle g_{1}(h)(u,v) =μj=12{Ω(huj)Δv¯j+(hv¯j)Δujdx\displaystyle=-\mu\sum_{j=1}^{2}\{\int_{\Omega}(h\cdot\nabla u_{j})\Delta\bar{v}_{j}+(h\cdot\nabla\bar{v}_{j})\Delta{u}_{j}\,\mathrm{d}x
SR(huj)(nv¯j)+(hv¯j)(nuj)\displaystyle-\int_{S_{R}}(h\cdot\nabla u_{j})(n\cdot\nabla\bar{v}_{j})+(h\cdot\nabla\bar{v}_{j})(n\cdot\nabla{u}_{j})
(hn)(v¯juj)ds}.\displaystyle-(h\cdot n)(\nabla\bar{v}_{j}\cdot\nabla u_{j})\,\mathrm{d}s\}. (5.45)

By (Δ+ω2)u=0(\Delta^{*}+\omega^{2})u=0 and integration by parts, we have

j=12\displaystyle\sum_{j=1}^{2} μΩ(hv¯j)Δujdx=(λ+μ)Ω(u)(hv¯)dx\displaystyle\mu\int_{\Omega}(h\cdot\nabla\bar{v}_{j})\Delta{u}_{j}\,\mathrm{d}x=(\lambda+\mu)\int_{\Omega}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})\,\mathrm{d}x
(λ+μ)SR(u)(n(hv¯))dsω2Ω(hv¯)udx.\displaystyle-(\lambda+\mu)\int_{S_{R}}(\nabla\cdot u)(n\cdot(h\cdot\nabla\bar{v}))\,\mathrm{d}s-\omega^{2}\int_{\Omega}(h\cdot\nabla\bar{v})\cdot u\,\mathrm{d}x. (5.46)

Integration by parts again gives

j=12μ\displaystyle\sum_{j=1}^{2}\mu Ω(huj)Δv¯jdx=\displaystyle\int_{\Omega}(h\cdot\nabla u_{j})\Delta\bar{v}_{j}\,\mathrm{d}x=
μΩ(hu):v¯dx+μSR(hu)(nv¯)ds.\displaystyle-\mu\int_{\Omega}\nabla(h\cdot\nabla u):\nabla\bar{v}\,\mathrm{d}x+\mu\int_{S_{R}}(h\cdot\nabla u)\cdot(n\cdot\nabla\bar{v})\,\mathrm{d}s. (5.47)

Combining (5.2)-(5.2) and (λ+μ)(u)n+μnu=0(\lambda+\mu)(\nabla\cdot u)n+\mu\partial_{n}u=0 on SRS_{R} gives

g1(h)(u,v)\displaystyle g_{1}(h)(u,v) =μΩ(hu):v¯dx(λ+μ)Ω(u)(hv¯)dx\displaystyle=\mu\int_{\Omega}\nabla(h\cdot\nabla u):\nabla\bar{v}\,\mathrm{d}x-(\lambda+\mu)\int_{\Omega}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})\mathrm{d}x
+ω2Ω(hv¯)udxμSR(hn)(u:v¯)ds.\displaystyle+\omega^{2}\int_{\Omega}(h\cdot\nabla\bar{v})\cdot u\,\mathrm{d}x-\mu\int_{S_{R}}(h\cdot n)(\nabla u:\nabla\bar{v})\,\mathrm{d}s. (5.48)

Apply the divergence theorem again, we obtain

Ω(h)(uv¯)+u(hv¯)dx\displaystyle\int_{\Omega}(\nabla\cdot h)(u\cdot\bar{v})+u\cdot(h\cdot\nabla\bar{v})\,\mathrm{d}x =Ω((uv¯)h)(hu)v¯dx\displaystyle=\int_{\Omega}\nabla\cdot((u\cdot\bar{v})h)-(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x
=SR(uv¯)(hn)dsΩ(hu)v¯dx.\displaystyle=\int_{S_{R}}(u\cdot\bar{v})(h\cdot n)\,\mathrm{d}s-\int_{\Omega}(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x. (5.49)

Combining (5.2)-(5.2) yields

g1\displaystyle g_{1} (h)(u,v)+g3(h)(u,v)=\displaystyle(h)(u,v)+g_{3}(h)(u,v)=
μΩ(hu):v¯dx(λ+μ)Ω(u)(hv¯)dx\displaystyle\mu\int_{\Omega}\nabla(h\cdot\nabla u):\nabla\bar{v}\,\mathrm{d}x-(\lambda+\mu)\int_{\Omega}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})\mathrm{d}x
ω2Ω(hu)v¯dx+ω2SR(uv¯)(hn)dsμSR(hn)(u:v¯)ds.\displaystyle-\omega^{2}\int_{\Omega}(h\cdot\nabla u)\cdot\bar{v}\,\mathrm{d}x+\omega^{2}\int_{S_{R}}(u\cdot\bar{v})(h\cdot n)\,\mathrm{d}s-\mu\int_{S_{R}}(h\cdot n)(\nabla u:\nabla\bar{v})\,\mathrm{d}s. (5.50)

For g2(h)(u,v)g_{2}(h)(u,v), similar discussion as (5.27)-(5.1) shows

g2(h)(u,v)\displaystyle g_{2}(h)(u,v) =(λ+μ)Ω(u)(hv¯)+(hu)(v¯)dx\displaystyle=(\lambda+\mu)\int_{\Omega}(\nabla\cdot u)\nabla\cdot(h\cdot\nabla\bar{v})+\nabla\cdot(h\cdot\nabla u)(\nabla\cdot\bar{v})\,\mathrm{d}x
(λ+μ)SR(u)(v¯)(nh)ds.\displaystyle-(\lambda+\mu)\int_{S_{R}}(\nabla\cdot u)(\nabla\cdot\bar{v})(n\cdot h)\,\mathrm{d}s. (5.51)

Notice uu^{*} satisfies (5.41)-(5.43), then

B(u,v)\displaystyle B(u^{*},v) =ω2SR(uv¯)(hn)ds\displaystyle=\omega^{2}\int_{S_{R}}(u\cdot\bar{v})(h\cdot n)\,\mathrm{d}s
(λ+μ)SR(u)(v¯)(nh)dsμSR(hn)(u:v¯)ds.\displaystyle-(\lambda+\mu)\int_{S_{R}}(\nabla\cdot u)(\nabla\cdot\bar{v})(n\cdot h)\,\mathrm{d}s-\mu\int_{S_{R}}(h\cdot n)(\nabla u:\nabla\bar{v})\,\mathrm{d}s. (5.52)

Combining wh|ΓR=0w_{h}|_{\Gamma_{R}}=0 and (5.2)-(5.2) gives

i=13gi(h)(u,v)=μΩ(h:u)v¯dx\displaystyle\sum^{3}_{i=1}g_{i}(h)(u,v)=\mu\int_{\Omega}\nabla(h:\nabla u)\nabla\bar{v}\,\mathrm{d}x
+(λ+μ)Ω(uh)(v¯)dx(λ+μ)SR(u)(v¯)(nh)ds\displaystyle+(\lambda+\mu)\int_{\Omega}\nabla\cdot(\nabla u\cdot h)(\nabla\cdot\bar{v})\,\mathrm{d}x-(\lambda+\mu)\int_{S_{R}}(\nabla\cdot u)(\nabla\cdot\bar{v})(n\cdot h)\,\mathrm{d}s
μSR(hn)(u:v¯)ds+ω2SR(uv¯)(hn)ds=B(wh+u,v).\displaystyle-\mu\int_{S_{R}}(h\cdot n)(\nabla u:\nabla\bar{v})\,\mathrm{d}s+\omega^{2}\int_{S_{R}}(u\cdot\bar{v})(h\cdot n)\,\mathrm{d}s=B(w_{h}+u^{*},v).

Then we arrive at

B(u~huuwh,v)=O(h1,2vH1(Ω)2)+O(u~huH1(Ω)2vH1(Ω)2h1,).B(\tilde{u}_{h}-u-u^{*}-w_{h},v)=O(\|h\|^{2}_{1,\infty}\|v\|_{H^{1}(\Omega)^{2}})+O(\|\tilde{u}_{h}-u\|_{H^{1}(\Omega)^{2}}\|v\|_{H^{1}(\Omega)^{2}}\|h\|_{1,\infty}).

to completes the proof. ∎

Remark. We can not get local stability from the Fre´\acute{\rm e}chet derivative like the Dirichlet problem. Let h(x)=kp(x)nh(x)=kp(x)n with k>0k>0 and p(x)C2(S,)p(x)\in C^{2}(S,\mathbb{R}). Assume p(x)0p(x)\not\equiv 0. For any vH1(Ω)2v\in H^{1}(\Omega)^{2}, the equality

(λ+μ)SRp(u)(v¯)dsμSRp(u:v¯)ds+ω2SRp(uv¯)ds=0-(\lambda+\mu)\int_{S_{R}}p(\nabla\cdot u)(\nabla\cdot\bar{v})\,\mathrm{d}s-\mu\int_{S_{R}}p(\nabla u:\nabla\bar{v})\,\mathrm{d}s+\omega^{2}\int_{S_{R}}p(u\cdot\bar{v})\,\mathrm{d}s=0

does not imply u|SR=0u|_{S_{R}}=0. In fact, we can see in [22] for electromagnetic scattering, the local stability holds only for lossy medium, i.e, non-real wave number.

6 Conclusion

In this paper, elastic cavity problem with Dirichlet or Neumann condition is reduced into a bounded domain by the DtN or NtD operator. Variational approaches is utilized to prove the uniqueness and existence of boundary value problem in a bounded domain. For inverse cavity problem, the Fre´\acute{\rm e}chet derivative is given for shape reconstruction and a local stability result is given for the Dirichlet problem. The stability results explicit with frequency for large elastic cavities like [4, 5] remains unsolved, which will be discussed in a forthcoming paper.

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